A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int\limits_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$

Well, can anyone prove this without using Residue theory? I actually thought of using the series representation of $\sin x$: $$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.

Michael Hardy
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    note that from $\int\limits_0^\infty \frac{\sin(x)}{x}dx=\frac{\pi}{2}$, we can get $\int_0^\infty\frac{\sin(x^n)}{x}dx=\frac{\pi}{2n}$ by a simple change of variables –  Nov 12 '12 at 02:04
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    Since no one has mentioned them yet, G.H. Hardy wrote two articles about approximately 12 different ways of doing this integral: in [1909](http://www.math.harvard.edu/~ctm/home/text/class/harvard/55b/10/html/home/hardy/sinx/sinx.pdf) and *Math. Gaz.* 8 (July 1916) pp. 301–303., although the latter is not very easy to find online. Both are available in his *Collected Works* and *The G. H. Hardy Reader*. – Chappers Feb 15 '17 at 23:56

32 Answers32


I believe this can also be solved using double integrals.

It is possible (if I remember correctly) to justify switching the order of integration to give the equality:

$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Notice that $$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$

This leads us to

$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Now the right hand side can be found easily, using integration by parts.

$$\begin{align*} I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\ &= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\ &= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}. \end{align*}$$ Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$ Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$

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    @Americo: I heard of this one a long time back from one of my teachers. I thought this will be well known, but I guess I could be mistaken about that. – Aryabhata Sep 23 '10 at 16:26
  • I believe this idea shows up in Melzak's Companion to Concrete Mathematics; I know he has a small stack of 'clever' ideas for integrals and I'm pretty sure this is one of them. – Steven Stadnicki Sep 24 '10 at 21:57
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    This is also the technique used in R. Durrett (2005), *Probability theory and examples*, 3rd ed., Duxbury, p. 470. It is Exercise 6.6 on that page. The justification of exchanging the order of integration actually comes from considering the strip $(0,a) \times (0,\infty)$ and observing that $\int_0^a \int_0^\infty |e^{-xy} \sin x| \,\mathrm{d} y\,\mathrm{d} x \leq a$, from whence Fubini's theorem can be applied. To get the result, we take $a \to \infty$. – cardinal Sep 18 '11 at 12:09
  • @Chandrasekhar: See this: http://math.stackexchange.com/questions/13344/proof-for-an-integral-involving-sinc-function – Aryabhata Feb 10 '12 at 18:14
  • @night owl: May I ask you why can the switch of integration happen, i.e., why is the first equation true? I thought this proof was based on Fubini's theorem. Thanks. – ShinyaSakai Jan 02 '13 at 08:51
  • @Aryabhata: I am sorry I @ ed the wrong person. Please read the previous comment. Thanks. – ShinyaSakai Jan 02 '13 at 08:56
  • @ShinyaSakai: I don't think we can apply Fubini's directly. – Aryabhata Jan 02 '13 at 18:09
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    this methods is elegant,and I computer the integral $\int e^{-xy}\sin x \text{dx}$: $$\begin{align*} I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\ &= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big) \end{align*} $$ – Laura Feb 04 '13 at 08:30
  • @Aryabhata You correctly say that $\int_{0}^a \int_{0}^\infty |e^{-xy}sin(x)| \le a$, and thus you can use Fubini, but what if a $\to \infty$? are you still allowed to use Fubini? My doubt comes from the fact that the you can't find a bound, since it would be $\infty$ Am i wrong? – Bman72 Jan 16 '14 at 14:33
  • @Ale: I believe cardinal said that... There might be different versions of Fubini (I am not aware of though...) – Aryabhata Jan 16 '14 at 22:08
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    @Ale: As regards your comment, yes, Fubini is applied correctly as stated in my comment. For each fixed $a$, the associated integral is bounded. Hence, Fubini can be used to compute the integral in two ways. Rearranging gives you a bound for $|\int_0^a \frac{\sin x}{x} - \frac{\pi}{2}|$ as a function of $a$. Then, take $a \to \infty$ to get the result. (Note that *some* argument like this is necessary since $\frac{\sin x}{x}$ is *not* integrable on $(0,\infty)$ in the Lebesgue sense.) Hope this helps. Cheers. :-) – cardinal Feb 17 '15 at 02:07
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    This method is basically the same as the usage of an inverse and a regular Laplace transform. The required substitution can be obtained systematically through contour integrals, but usually the point of using it to avoid complex integration... – 1010011010 Dec 23 '15 at 20:53
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    How to justify the interchange in the order of integration could bear examination. Note for example that $$ \int_0^1 \int_0^1 \frac{x^2-y^2} {(x^2+y^2)^2} \, dx \, dy \ne \int_0^1 \int_0^1 \frac{x^2-y^2} {(x^2+y^2)^2} \, dy \, dx. $$ It would be enough to prove that $ \displaystyle \iint\limits_{[0,\infty)^2} |e^{-xy} \sin x| \, d(x,y) < +\infty. $ But is that true? $\qquad$ – Michael Hardy Jul 25 '17 at 06:18
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    @MichaelHardy: See cardinal's comments (note: multiple) in this comment thread. Caveat: I haven't gone through them carefully, but they seem to give a method of justifying the order switch. – Aryabhata Jul 25 '17 at 17:14
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    ok, I see that "cardinal" addressed this issue. It's mildly subtle in that you can't just work directly on $[0,\infty)^2$ but need to work on $[0,a) \times [0,\infty)$ and then afterwards take a limit as $a\to\infty. \qquad$ – Michael Hardy Jul 25 '17 at 20:03
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    This method, including the issues raised by me and by "cardinal" and barely hinted at in the answer itself, appears in Walter Rudin's _Real and Complex Analysis_ as an exercise. – Michael Hardy Jul 25 '17 at 20:07
  • This is similar to Feynman's method. – Peter Driscoll Aug 21 '17 at 04:07

Here's another way of finishing off Derek's argument. He proves $$\int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2.$$ Let $$I_n=\int_0^{\pi/2}\frac{\sin(2n+1)x}{x}dx= \int_0^{(2n+1)\pi/2}\frac{\sin x}{x}dx.$$ Let $$D_n=\frac\pi2-I_n=\int_0^{\pi/2}f(x)\sin(2n+1)x\ dx$$ where $$f(x)=\frac1{\sin x}-\frac1x.$$ We need the fact that if we define $f(0)=0$ then $f$ has a continuous derivative on the interval $[0,\pi/2]$. Integration by parts yields $$D_n=\frac1{2n+1}\int_0^{\pi/2}f'(x)\cos(2n+1)x\ dx=O(1/n).$$ Hence $I_n\to\pi/2$ and we conclude that $$\int_0^\infty\frac{\sin x}{x}dx=\lim_{n\to\infty}I_n=\frac\pi2.$$

Robin Chapman
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Here's one more, just for the fun of it. For $\theta$ not an integer multiple of $2 \pi$, we have $$\sum \frac{e^{i n \theta}}{n} = -\log(1-e^{i \theta}).$$ Taking imaginary parts, for $0 < \theta < \pi$, we have $$\sum \frac{\sin (n \theta)}{n} = -\mathrm{arg}(1-e^{i \theta}) = \pi/2-\frac{\theta}{2}.$$ Draw the isosceles triangle with vertices at $0$, $1$ and $e^{i \theta}$ to see the second equality.

So $\displaystyle \sum \theta \cdot \frac{\sin (n \theta)}{n \theta} = \pi/2-\frac{\theta}{2}$. The right hand side is a right-hand Riemann sum for $\int \frac{\sin t}{t} dt$, with intervals of width $\theta$. So, taking the limit as $\theta \to 0$, we get $$\int\limits_0^\infty \frac{\sin t}{t} dt=\frac{\pi}{2}$$.

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David E Speyer
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    Sorry for digging out a 4 years old post, but how does one justify that the limit of the sum is actually the integral we are interested in? I only know that this kind of Riemann sums works for bounded intervals and I'm not convinced this works for improper integrals. Is there a general result which justifies this? – Wojowu Mar 26 '16 at 20:14

One easiest way to get this integral is to evaluate the following improper integral with parameter $a$: $$ I(a)=\int_0^\infty e^{-ax}\frac{\sin x}{x}dx, a\ge 0.$$ It is easy to see $$I'(a)=-\int_0^\infty e^{-ax}\sin xdx=\frac{e^{-ax}}{a^2+1}(a\sin x+\cos x)\big|_0^\infty=-\frac{1}{a^2+1}.$$ Thus $$I(\infty)-I(0)=-\int_0^\infty\frac{1}{a^2+1}da=-\frac{\pi}{2}.$$ Note $I(\infty)=0$ and hence $I(0)=\frac{\pi}{2}$.

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    +1 This is esentially the Feynman method mentioned in "Chris's sis" answer; in the linked pdf there a justification for deriving under the integral is provided. – leonbloy Aug 10 '13 at 00:48
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    Why does $I(\infty) = 0$? Instead of $I(\infty) - I(0) = ...$, shouldn't it be $I(\infty) - \lim_{ a \rightarrow 0^{+} }I(a) = ...$ because you haven't shown $I(0)$ converges? So it still remains to show $\lim_{ a \rightarrow 0^{+} }I(a) = I(0)$? – LucasSilva May 20 '15 at 05:53
  • @LucasSilva, I omitted the details. – xpaul May 20 '15 at 15:22
  • @xpaul: I've worked out why $I(\infty) = 0$: $|e^{-ax} \sin x| \leq e^{-x}$ for $a \geq 1$, so the integrand of $I(a)$ is $L^1$ for large $a$, and we can use the dominated convergence theorem. – LucasSilva May 20 '15 at 15:35
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    @xpaul : The issue with $I(0)$ is still unresolved. The issue also exists on the wikipedia page: http://en.wikipedia.org/wiki/Dirichlet_integral#Differentiation_under_the_integral_sign – LucasSilva May 20 '15 at 15:37
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    The issue also exists in the link referred to by @leonbloy : http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf – LucasSilva May 20 '15 at 16:28
  • $I(0)$ exists because integration by parts gives $\lim_{t \rightarrow \infty} \int_{\pi/2}^{t} \frac{\sin x}{x} dx = - \lim_{t \rightarrow \infty} \int_{\pi/2}^{t} \frac{\cos x}{x^2} dx$. The improper integral on the right-hand side exists because $\int_{\pi/2}^{t} |\frac{\cos x}{x^2}| dx$ is bounded by $\int_{\pi/2}^{\infty} \frac{1}{x^2}$ as $t \rightarrow \infty$. See Theorem 10.33 of Apostol's "Mathematical Analysis" or http://en.wikipedia.org/wiki/Improper_integral#Improper_Riemann_integrals_and_Lebesgue_integrals – LucasSilva May 20 '15 at 16:53
  • [YouTube video](https://www.youtube.com/watch?v=s1zhYD4x6mY) – md2perpe Jun 24 '20 at 22:09

Here is a sketch of another elementary solution based on a proof in Bromwich's Theory of Infinite Series.

Using $\sin(2k+1)x-\sin(2k-1)x = 2\cos2kx\sin x$ and summing from k=1 to k=n we have $$\sin(2n+1)x = \sin x \left( 1+ 2 \sum_{k=1}^n \cos 2kx \right),$$

and hence $$ \int_0^{\pi/2} {\sin(2n+1)x \over \sin x} dx = \pi/2. \qquad (1)$$

Let $y=(2n+1)x$ and this becomes $$ \int_0^{(2n+1)\pi/2} {\sin y \over (2n+1) \sin (y/(2n+1))} dy = \pi/2.$$

and since $\lim_{n \to \infty} (2n+1) \sin { y \over 2n+1} = y$ it suggests that there is a proof lurking in there somewhere.

So let's put $$\begin{align} I_n &= \int_0^{n\pi/(2n+1)} {\sin(2n+1)x \over \sin x} dx \ &= \sum_{k=0}^{n-1} \int_{k\pi/(2n+1)}^{(k+1)\pi/(2n+1)} {\sin(2n+1)x \over \sin x} dx. \end{align}$$

Hence we have $I_n = u_0 – u_1 + u_2 \cdots + (-1)^{n-1}u_{n-1},$ where $u_k$ is a decreasing sequence of positive terms. We can see this from the shape of the curve $y = \sin(2n+1)x / \sin x,$ which crosses the x-axis at $\pi/(2n+1), 2\pi/(2n+1),\ldots,n\pi/(2n+1).$ (I said that this is just a sketch, you have to check the details.)

Hence the sequence $I_n$ converges, and by (1) it converges to $\pi/2.$

Now if we make the substitution $y=(2n+1)x$ we see that $$u_k = \int_{k\pi}^{(k+1)\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy,$$

and since $I_n$ can be written as an alternating sequence of decreasing positive terms we can truncate the sequence wherever we like and the value of $I_n$ lies between two successive partial summations. Hence

$$ \int_{0}^{2m\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy < I_n < \int_{0}^{(2m+1)\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy. \qquad (2)$$

for any m such that $2m+1 \le n.$ (Take $m=[\sqrt{n}],$ say, $n \ge 6.$)

Now $$\left| { \sin y \over y} - {\sin y \over (2n+1) \sin(y/(2n+1))} \right| < { \pi^2(2m+1)^2 \over 3(2n+1)^2}$$ and so this difference tends to zero uniformly in the interval $0 \le y \le (2m+1)\pi$ and so by taking the $\lim_{n \to \infty}$ in (2) we obtain $$\int_0^{\infty} { \sin x \over x } dx = { \pi \over 2}.$$

Derek Jennings
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Let's consider the integrals
$$I_1(t)=\int_t^{\infty}\frac{\sin(x-t)}{x}dx\qquad\mbox{ and }\qquad I_2(t)=\int_0^{\infty}\frac{e^{-tx}}{1+x^2}dx,\qquad t\geq 0.$$ A direct calculation shows that $I_1(t)$ and $I_2(t)$ satisfy the ordinary differential equation $$y''+y=\frac{1}{t},\qquad t>0.$$ Therefore, the difference $I(t)=I_1(t)-I_2(t)$ satisfy the homogeneous differential equation $$y''+y=0,\qquad t>0,$$ hence it should be of the form $$I(t)=A\sin (t+B) $$ with some constants $A$, $B$. But $I_1(t)$ and $I_2(t)$ both converge to $0$ as $t\to\infty$. This implies that $A=0$ and $I_1(t)=I_2(t)$ for all $t\geq 0$. Finally, we have that $$\int_0^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\lim_{n\to\infty}\left(\arctan(n)\right)-\arctan(0)=\frac{\pi}{2}.$$

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Andrey Rekalo
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Maybe I'm flogging a dead horse, but nobody has mentioned the standard suspiciously circular (see the comments) Fourier analytic proof yet:

Let $f(t)=1$ for $|t|<1$ and 0 otherwise. Then the Fourier transform is $$ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt = \int_{-1}^{1} e^{-i\omega t} dt = \frac{e^{-i\omega} - e^{i\omega}}{-i\omega} = \frac{2\sin\omega}{\omega}.$$

Fourier's inversion formula states that $$ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} d\omega $$ if $f$ is (say) differentiable at $t$. In our case, we get in particular that $$ 1 = f(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{2\sin\omega}{\omega} d\omega = \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin\omega}{\omega} d\omega. $$

(EDIT: Even if this is not really a proof, it's still a good thing to be aware of, since one can use similar ideas to integrate powers of $\sin\omega/\omega$, or integrals like these.)

Hans Lundmark
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    Often proofs of the Fourier inversion theorem *use* the value of the sine integral. Certainly the one I learned as an undergraduate did. To me this is reminiscent of the argument that $\lim_{x\to0}(\sin x)/x=1$ *by L'Hopital's rule* :-( – Robin Chapman Oct 13 '10 at 10:14
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    @Robin Chapman: Hmm, that's true. Very good point. Maybe that's why nobody gave this answer! PS. You need to get rid of the reflex to hit the Return key before you're done writing your comments. :) – Hans Lundmark Oct 13 '10 at 10:33
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    +1, because posts like these make me want to *properly* learn fourier analysis as soon as possible. Ps. the proof of the inversion formula in Rudin's book doesn't get anywhere near of making use of the value in this integral, as far as I can remeber. – Sam Apr 22 '11 at 14:00
  • @RobinChapman Can you point me to a reference to a proof the Fourier inversion theorem that uses the value of the sine integral? – LucasSilva May 20 '15 at 17:26
  • @LucasSilva: Robin hasn't been active on this site for several years (unfortunately). – Hans Lundmark May 20 '15 at 20:05
  • @HansLundmark Thanks for letting me know. Maybe someone else will point me in the right direction. – LucasSilva May 20 '15 at 20:31
  • Not sure why fourier inversion is valid. $F$ is not of $L^1$. – Rubertos May 05 '16 at 03:05
  • @Rubertos: https://en.wikipedia.org/wiki/Fourier_inversion_theorem#Integrable_functions_in_one_dimension – Hans Lundmark May 05 '16 at 11:44
  • To apply fourier inversion formula, one must have $f,\hat f\in L^1$. But $F$ is not of $L^1$, so inversion formula is not valid. You cannot simply use that. Nevertheless, $F$ is in $L^2$. Plancherel theorem then implies that $\frac{1}{2\pi} \int_{-N}^N F(w)e^{iwt} dw \to f(t)$ in $L^2$ norm as $N\to\infty$. Thus there exists a subsequence $\{N_k\}$ such that $\frac{1}{2\pi} \int_{-N_k}^{N_k} F(w)e^{iwt} dw \to f(t)$ pointwise a.e. – Rubertos May 05 '16 at 12:46
  • @HansLundmark I'm asking because I'm curious.. Am I thinking something wrong? Why $f$ differentjable at $t$ makes the Fourier inversion formula valid? Don't we need details I wrote in the above comment? – Rubertos May 05 '16 at 12:47
  • There are several variants of the inversion formula. The one that I linked to only requires that $f$ belongs to $L^1$ and is piecewise smooth (but you have to take the Fourier inversion integral in the sense $\lim_{r\to\infty}\int_{-r}^r$). It's Theorem 7.6 in Folland's *Fourier Analysis and Its Applications*. – Hans Lundmark May 05 '16 at 13:57
  • @Sam, which Rudin book is this derivation found? –  Jan 17 '17 at 03:05
  • @Zermelo's_Choice: *Real and Complex Analysis*, chapter 9. – Hans Lundmark Jan 17 '17 at 08:26

\begin{align} \int_{-\infty}^{\infty}{\sin(x) \over x} \,{\rm d}x & = \int_{-\infty}^{\infty} \left({1 \over 2}\,\int_{-1}^{1} {\rm e}^{{\rm i}kx}\,{\rm d}k\right) \,{\rm d}x \\[5mm] & = \pi\int_{-1}^{1}\ \int_{-\infty}^{\infty}{\rm e}^{{\rm i}kx}\,{{\rm d}x \over 2\pi}\,\,{\rm d}k \\[5mm] & = \pi\int_{-1}^{1}\delta(k)\,{\rm d}k = \pi \end{align}

Felix Marin
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I evaluated this integral in this answer where I started with $$ \begin{align} \sum_{k=1}^\infty\frac{\sin(2kx)}{k} &=\sum_{k=1}^\infty\frac{e^{i2kx}-e^{-i2kx}}{2ik}\\ &=\frac1{2i}\left(-\log(1-e^{i2x})+\log(1-e^{-i2x})\right)\\ &=\frac1{2i}\log(-e^{-i2x})\\[4pt] &=\frac\pi2-x\quad\text{for }x\in\left(0,\pi\right)\tag{1} \end{align} $$ Setting $x=\frac a2$, we get $$ \sum_{k\in\mathbb{Z}}\frac{\sin(ka)}{ka}=\frac\pi a\tag{2} $$ where we set $\frac{\sin(ka)}{ka}=1$ when $k=0$. Multiplying $(2)$ by $a$ and setting $a=\frac1n$ yields $$ \sum_{k\in\mathbb{Z}}\frac{\sin(k/n)}{k/n}\frac1n=\pi\tag{3} $$ and $(3)$ is a Riemann Sum for the integral $$ \int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\pi\tag{4} $$

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Note: Laplace transforms, $$\int_{0}^{\infty}e^{-st}f(t)dt=L[f(t)]$$ $$L\left[\frac{f(t)}{t}\right]=\int_{s}^{\infty}L[f(t)]\ ds$$ & $$L[\sin t]=\frac{1}{1+s^2}$$ Now, we have $$\int_{0}^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}e^{-(0)x} \frac{\sin x}{x}\ dx$$$$=L\left[\frac{\sin x}{x}\right]_{s=0}$$ $$=\int_{s=0}^{\infty}L\left[\sin x\right]\ ds$$ $$=\int_{s=0}^{\infty}\frac{1}{1+s^2}\ ds$$ $$=[\tan^{-1}(s)]_{0}^{\infty}$$$$=\tan^{-1}(\infty)-\tan^{-1}(0)$$$$=\frac{\pi}{2}$$

jeanne clement
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Harish Chandra Rajpoot
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    I used Laplace Transforms as well, but in a different manner https://math.stackexchange.com/a/2952227/150203 –  Oct 12 '18 at 04:17

These proofs looked very intriguing the multiple ways to go about the same problem. I looked up toward the ceiling and then it dawned on me that there was another way to do this with this particular function as follows:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$The method of attack of use would be Laplace Transforms


$$ \lim_{t \to 0} ~ \dfrac{f(t)}{t} ~;~ \text{exist and is a finite number.}$$

$${\cal L} \left\{ \frac{\sin(t)}{t} \right\}=\int_0^\infty \! {\cal L} \left\{ \sin(t) \right\} ~ \mathrm{d} \sigma=\int_0^\infty\! \frac{1}{\sigma^2+1} \mathrm{d} \sigma=\tan^{-1}(\sigma) ~ {\LARGE|_{\sigma=0}^{\sigma=\infty}}=\frac{\pi}{2}- \arctan(0)$$

So we see that we get the result of: $\dfrac{\pi}{2}~~~$ $\Big(\because~\arctan(0)=0 ~\Big)$.

Michael Hardy
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night owl
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    How is $\cal{L}\{\frac{\sin(t)}{t}\}=\int^{\infty}_0 \cal{L}\{\sin(t)\}d\sigma$? And also, how did you get a $\sigma$ there? – Aditya Agarwal Dec 30 '15 at 09:34

This one I found in The American Mathematical Monthly from 1951 in the article 'A simple evaluation of an improper integral' written by Waclaw Kozakiewicz.

Theorem (Riemann). If $f(x)$ is Riemann integrable in the interval $a \leq x \leq b$, then: $$\lim_{k \to +\infty} \int_a^b f(x) \sin kx \; dx = 0 \;.$$

Next, notice that: $$\int_0^\pi \frac{\sin \left(n+\frac{1}{2}\right)x}{2 \sin \frac{x}{2}}\; dx = \frac{\pi}{2} \; ,n = 0,1,2,\ldots \quad (1)$$ and let: $$\phi(x) = \begin{cases} 0 & , \;x = 0 \\ \frac{1}{x} - \frac{1}{2 \sin \frac{x}{2}} =\frac{2 \sin \frac{x}{2} - x}{2x \sin \frac{x}{2}} & ,\; 0 < x \leq \pi \; . \end{cases}$$ Then $\phi(x)$ is continuous and satisfies Riemann theorem, so choosing $k = n + \frac{1}{2}$ we write: $$\lim_{n \to +\infty}\int_0^{\pi} \left(\frac{1}{x} - \frac{1}{2 \sin \frac{x}{2}} \right) \sin \left(n+\frac{1}{2}\right)x \; dx = 0 \;.$$ But taking $(1)$ into account we have: $$\lim_{n \to +\infty} \int_0^\pi \frac{\sin \left(n+\frac{1}{2}\right)x}{x} \; dx = \frac{\pi}{2}\;.$$ Using substitution $u = \left(n+\frac{1}{2}\right)x$ and knowing that $\int_0^{+\infty} \frac{\sin x}{x} \; dx$ converges we finally have:

$$\int_0^{+\infty} \frac{\sin x}{x} \; dx = \lim_{n \to +\infty} \int_0^{\left(n+\frac{1}{2}\right)\pi}\frac{\sin u}{u} \; du = \frac{\pi}{2}\;.$$

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We can decompose interval $[0,+\infty)$ into intervals of length $\frac{\pi}{2}$. Then we'll have:

$$I = \int_0^{+\infty} \frac{\sin x}{x} \,dx = \sum_{n=0}^{+\infty} \int_{n\pi / 2}^{(n+1)\pi / 2} \frac{\sin x}{x} \,dx$$ Now consider the case when $n$ is even i.e. $n=2k$ and substitute $x = k\pi + t$:

$$\int_{2k\pi /2}^{(2k+1)\pi / 2} \frac{\sin x}{x} \,dx = (-1)^k \int_0^{\pi/ 2} \frac{\sin t}{k\pi + t} \, dt$$

and for odd $n$ we have $n=2k-1$ and we use substitution $x = k\pi-t$:

$$\int_{(2k-1)\pi /2}^{2k \pi / 2} \frac{\sin x}{x} \,dx = (-1)^{k-1} \int_0^{\pi/ 2} \frac{\sin t}{k\pi - t} \, dt$$

Hence we obtain:

$$I = \int_0^{\frac{\pi}{2}} \sin t \cdot \left[ \frac{1}{t} + \sum_{k = 1}^{+\infty} (-1)^k \left( \frac{1}{t+k\pi} + \frac{1}{t-k\pi} \right) \right] \, dt$$ But in square bracket we have expansion of $\frac{1}{\sin x}$ into partial fractions, hence the result follows: $$I = \int_0^{\frac{\pi}{2}} dt = \frac{\pi}{2}$$

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  • (+1) I like your using the partial fraction expansion of the cosecant function. And it is easy to justify the legitimacy of interchanging the integral and series. – Mark Viola May 09 '20 at 23:54

See http://en.wikipedia.org/wiki/Dirichlet_integral for a proof using differentiation under the integral sign.

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I got linked to this old question from a more recent one, and I hope that you don't mind me adding a somewhat bizarre way of doing calculating this integral, using Bessel functions.

I'm aware of that this way is not the shortest way of obtaining the result, and the facts I give on Bessel functions are standard, and can be found in (probably) any book on Bessel functions. Therefore, some details will be left to be checked by the interested reader.

I have never seen this way to calculate the integral $\int_0^{+\infty}\frac{\sin x}{x}\,dx$, but I claim no originality. If someone has seen it, please tell in a comment. Here it goes:

Let us define the $n$th Bessel function $J_n$ by the integral $$ J_n(x)=\frac{1}{\pi}\int_0^\pi \cos(n\theta-x\sin \theta)\,d\theta. $$ We will only work with the cases $n=0$ and $n=1$. The function $J_n$ solves the Bessel differential equation $$ x^2y''(x)+xy'(x)+(x^2-n^2)y(x)=0, $$ and moreover $D J_0(x)=-J_1(x)$ and $D(xJ_1(x))=xJ_0(x)$.

Our first statement is that $$ \frac{\sin x}{x}=\int_0^1\frac{y J_0(yx)}{\sqrt{1-y^2}}\,dy.\tag{1} $$ Indeed, define $f$ as $$ f(x)=\int_0^1\frac{xy J_0(yx)}{\sqrt{1-y^2}}\,dy $$ Then $$ f'(x)=\int_0^1 \frac{yJ_0(xy)-xy^2 J_1(xy)}{\sqrt{1-y^2}}\,dy $$ and $$ f''(x)=-\int_0^1 \frac{xy^3J_0(xy)+y^2J_1(xy)}{\sqrt{1-y^2}}\,dy, $$ and so $$ f''(x)+f(x)=\int_0^1 xy\sqrt{1-y^2}J_0(xy)-\frac{y^2}{\sqrt{1-y^2}}J_1(xy)\,dy=0. $$ (Here we integrated by parts in the last step.) Moreover, $$ f(0)=0,\quad\text{and}\quad f'(0)=\int_0^1 \frac{y}{\sqrt{1-y^2}}\,dy =1. $$ Thus $f(x)=\sin x$ and the equality (1) follows. Thus, we can write $$ \int_0^{+\infty} \frac{\sin x}{x}\, dx = \int_0^{+\infty}\int_0^1 \frac{y J_0(yx)}{\sqrt{1-y^2}}\,dy\, dx. $$ Next, we change the order of integration, and use the integral $$ \int_0^{+\infty} J_0(u)\,du=1,\tag{2} $$ to find that $$ \int_0^{+\infty} \frac{\sin x}{x}\, dx = \int_0^1 \frac{1}{\sqrt{1-y^2}}\,dy=\arcsin 1-\arcsin 0=\frac{\pi}{2}. $$ I remains to prove (2), which certainly follows by letting $x\to 0^+$ in $$ \int_0^{+\infty} J_0(u)e^{-xu}\,du=\frac{1}{\sqrt{1+x^2}}. $$ This is just the Laplace transform of $J_0$, and one can use the representation $J_0(u)=\frac{2}{\pi}\int_0^{\pi/2} \cos(u\cos\theta)\,d\theta$ to obtain it, $$ \begin{aligned} \int_0^{+\infty}J_0(u)e^{-xu}\,du &= \int_0^{+\infty} e^{-xu}\frac{2}{\pi}\int_0^{\pi/2}\cos(u\cos\theta)\,d\theta\\ &=\frac{2}{\pi}\int_0^{\pi/2} \int_0^{+\infty}e^{-xu}\cos(u\cos\theta)\,du\,d\theta\\ &=\frac{2}{\pi}\int_0^{\pi/2}\frac{x}{x^2+\cos^2\theta}\,d\theta\\ &=\frac{2}{\pi}\biggl[\frac{\arctan\Bigl(\frac{x\tan \theta}{\sqrt{1+x^2}}\Bigr)}{\sqrt{1+x^2}}\biggr]_0^{\pi/2}\\ &=\frac{1}{\sqrt{1+x^2}}. \end{aligned} $$

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    Just the other day I stumbled upon the fact that $\frac{\sin x}{x}=\int_0^1\frac{y J_0(yx)}{\sqrt{1-y^2}}\,dy $ when I was playing around with the [Abel transform](http://mathworld.wolfram.com/AbelTransform.html), and I immediately wondered if anyone had ever thought of using it to evaluate the Dirichlet integral. (+1) – Random Variable May 13 '17 at 02:15
  • @RandomVariable How did you prove that equality? – Poltroon Feb 28 '21 at 03:17
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    @Poltroon I must have used the integral representation $J_{0}(x) = \frac{2}{\pi} \int_{0}^{1} \frac{\cos (xt)}{\sqrt{1-t^{2}}} \, \mathrm dt $, but I don't remember how. Using the second form of the Abel transform and the alternative integral representation $J_{0}(x) = \frac{2}{\pi} \int_{1}^{\infty} \frac{\sin(xt)}{\sqrt{t^{2}-1}} \, \mathrm dt $, I get $\frac{\cos x}{x} = \int_{1}^{\infty} \frac{y J_{0}(xy)}{\sqrt{y^{2}-1}} \, \mathrm dy$, which is listed on that MathWorld page. – Random Variable Mar 01 '21 at 14:53

I'd add here the Feynman way, a very powerful, elegant and fast method to work out such things. You find here the example from $-\infty$ to $\infty$, but since the integrand is even, by dividing the result by 2 we get our required result.


user 1591719
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Another iteration of this question came up, and I have an answer that isn't currently here. So I present yet another solution.

We want to show that $\int_{0} ^{\infty} \frac{\sin x }{x} \mathrm{d}x = \pi/2.$

First, let's show that it converges. We let $I_{ab} = \int_a^b \frac{\sin x}{x}$, and consider the limits $a \to 0, b \to \infty$. $a \to 0$ is easy, so we don't worry about it. $\frac{\sin x}{x}$ is continuous on this domain, so all we really want is for the upper limit to behave nicely.

Note that $I_{ab} = \int \frac{\sin x}{x} = \int \frac{1}{x} \frac{\mathrm{d} (1 - \cos x)}{\mathrm{d} x}$, and so we can use integration by parts. We then get

$$I_{ab} = \frac{1 - \cos b}{b} - \frac{1 - \cos a}{a} + \int_a^b \frac{1 - \cos x}{x^2}$$

This clearly converges. In fact, one can see that both $\cos$ terms disappear in the limit. It's more important to simply note that the integral converges.

Knowing that, we continue the trend of the other answers and show that $\displaystyle \int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2$

We show the following: $$1 + 2 \cos 2t + 2 \cos 4t + \ldots + 2 \cos 2nt = \frac{\sin(2n + 1)t}{\sin t}$$

We do this with $\sin a - \sin b = 2 \sin(\frac{a-b}{2}) \cos(\frac{a + b}{2})$, so that we also get $\sin(2k + 1)t - \sin(2k -1)t = 2\sin(t) \cos (2kt)$. Thus $1 + 2 \cos 2t + \ldots + 2 \cos 2nt = 1 + \frac{1}{\sin t} \left[ \sum \sin(2k+1)t - \sin(sk-1)t \right] $

$\phantom{1 + 2 \cos 2t + \ldots + 2 \cos 2nt} = 1 + \frac{1}{\sin t} [\sin(2n + 1)t - \sin t]$

$\phantom{1 + 2 \cos 2t + \ldots + 2 \cos 2nt} = \frac{\sin(2n + 1)t}{\sin t}$

We did this just so that we could then say that

$$\int_0^{\pi/2} \frac{\sin (2n + 1)t}{\sin t} = \int_0 ^{\pi /2} (1 + 2 \cos 2t + 2 \cos 4t + \ldots + 2 \cos 2nt) = $$

$$\phantom{\frac{\sin (2n + 1)t}{\sin t}} = \frac{\pi}{2} + \left[ \sin 2t + \frac{\sin 4t}{2} + \ldots + \frac{\sin 2nt }{n}\right]_0^{\pi/2} = \frac{\pi}{2}$$

And thus we have it.

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    You really helped me with this. I'm a math freshman and currently have only studied basic academic math courses. This really helped me proof this for my homework. :) – Ory Band May 05 '12 at 16:02
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    Aren't there a few $dx$s and $dt$s missing? – JMCF125 Apr 05 '14 at 12:26

Three (more or less) elementary proofs in one answer.

We may notice that: $$ \int_{-\infty}^{+\infty}\frac{\sin x}{x}\,dx = \int_{-\pi/2}^{\pi/2}\sin(x)\left(\frac{1}{x}+\sum_{m\geq 1}\frac{(-1)^m 2x}{x^2-m^2\pi^2}\right)\,dx \tag{1}$$ since $\sin(x+\pi)=-\sin(x)$. We may study the singularities of $$ \frac{1}{x}+\sum_{m\geq 1}\frac{(-1)^m 2x}{x^2-m^2\pi^2} = \sum_{m\in\mathbb{Z}}\frac{(-1)^m}{x-m\pi}\tag{2}$$ to deduce it is exactly $\frac{1}{\sin x}$, so the RHS of $(1)$ simply equals $\color{red}{\pi}$. Or we may notice that $$ \forall \alpha>0,\qquad \int_{0}^{+\infty}\frac{\sin(\pi \alpha x)}{x}\,dx = C\tag{3}$$ and consider the Fourier series of a sawtooth-wave, divided by $x$: $$ f(x) = \sum_{n\geq 1}\frac{2(-1)^{n+1}\sin(\pi nx)}{\pi nx}. \tag{4}$$ By $(3)$, $\int_{0}^{+\infty}f(x)\,dx$ equals $\frac{2C}{\pi}\log 2$. On the other hand $x\,f(x)$ is piecewise linear, hence: $$\begin{eqnarray*}\int_{0}^{+\infty}f(x)\,dx &=& \int_{0}^{1}\frac{x}{x}\,dx+\int_{1}^{3}\frac{x-2}{x}\,dx+\int_{3}^{5}\frac{x-4}{x}\,dx+\ldots \\&=&1+\sum_{k\geq 1}\left(2-2k\log\frac{2k+1}{2k-1}\right)\tag{5}\end{eqnarray*}$$ and by summation by parts: $$ \sum_{k=1}^{N}2k\log\frac{2k+1}{2k-1} = 2N\log(2N+1)-2\sum_{k=1}^{N-1}\log(2k+1)\\=2N\log(2N+1)-2\log((2N-1)!!)\tag{6} $$ so $\int_{0}^{+\infty}f(x)\,dx=\log 2$, then $\color{red}{C=\frac{\pi}{2}}$, follow from Stirling's approximation. In order to compute $$ 2iC=\int_{0}^{+\infty}\frac{e^{ix}-e^{-ix}}{x}\,dx\tag{6}$$ we may also use the complex version of Frullani's theorem, leading to: $$ 2iC = \lim_{\varepsilon\to 0^+}\text{Log}\left(\frac{i+\varepsilon}{-i+\varepsilon}\right)=\pi i.\tag{7} $$

It is interesting to point out that $(5)$ gives a nice by-product, i.e. $$\begin{eqnarray*} \log(2) = 1+\sum_{k\geq 1}\left(2-2k\cdot 2\,\text{arctanh}\left(\frac{1}{2k}\right)\right) &=& 1-\sum_{k\geq 1}\sum_{n\geq 1}\frac{2}{(2n+1)(2k)^{2n}}\\&=&1-\sum_{n\geq 1}\frac{2\,\zeta(2n)}{(2n+1)4^n}\tag{8}\end{eqnarray*}$$ that can also be derived from: $$ \pi x \cot(\pi x)=1-\sum_{n\geq 1}2x^{2n}\zeta(2n)\tag{9} $$ by integrating both sides over the interval $\left(0,\frac{1}{2}\right)$: $$ \int_{0}^{1/2}\pi x\cot(\pi x)\,dx=\frac{1}{\pi}\int_{0}^{\pi/2}\frac{z\cos(z)}{\sin z}\,dz \stackrel{IBP}{=} -\frac{1}{\pi}\int_{0}^{\pi/2}\log(\sin z)\,dz\tag{10} $$ where the last integral can be computed through Riemann sums (!!!) since $$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n\pi}\right)=\frac{2n}{2^n}.\tag{11}$$

Jack D'Aurizio
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  • proving the partial fraction decomposition of $\frac{1}{\sin(z)}$ is supposed to be elementary :) ? – reuns Jul 03 '16 at 14:01
  • @user1952009: well, it is not terribly difficult to show that $\frac{1}{\sin z}$ has simple poles at $\pi\mathbb{Z}$ and the residues match the residues of $\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{z-\pi n}$ at the same points. – Jack D'Aurizio Jul 03 '16 at 14:06
  • @user1952009: anyway, it can also be derived from the Weierstrass product of the sine function through logarithmic differentiation. – Jack D'Aurizio Jul 03 '16 at 14:07
  • yes, the only way I know for all this is proving $\frac{1}{\sin(x)}$ or $\frac{1}{e^x-1}$ or $\tan(x)$ minus their poles are bounded entire functions, so by Liouville it reduces to a constant, and for $1/\sin(x)$ the constant should be easy to find – reuns Jul 03 '16 at 14:18
  • @user1952009: in this case we are lucky since $\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{z-n\pi}$ is not difficult to evaluate at, say, $z=\pi/2$. – Jack D'Aurizio Jul 03 '16 at 14:21
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    oh ok got it $\sum_{n=-N}^N \frac{(-1)^n}{2n-1} = 1+ \frac{(-1)^n}{-2N-1}$, tks didn't notice – reuns Jul 03 '16 at 14:30

In the book Advanced Calculus by Angus Taylor it is shown that, if $a\gt 0$,

$$\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt=\arctan\dfrac{x}{a}.\tag{1}$$

If $x>0$,

$$\displaystyle\int_0^{\infty}\dfrac{\sin xt}{t}dt=\dfrac{\pi}{2}\tag{2}$$

follows from $(1)$, observing that the integrand is $G(0)$ for

$$G(a)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt,\tag{3}$$

$G$ is uniformly convergent when $a\ge 0$, and $G(a)$ approaches $G(0)$ as $a$ tends to $0^+$.

Answer to Qiaochu: $(1)$ is proved as an application of the following theorem [Angus Taylor, Advanced Caluculus, p. 668] to $$F(x)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt.$$

Let $$F(x)=\displaystyle\int_c^{\infty}f(t,x)dt$$ be convergent when $a\le x\le b$. Let $\dfrac{\partial f}{\partial x}$ be continuous in $t,x$ when $c\le t,a\le x\le b$, and let $\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt$ converge uniformly on $[a,b]$. Then $$F'(x)=\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt.$$

Américo Tavares
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I think we should have Euler's original proof (from E675, translation available here). We start with $$ \int_0^{\infty} x^{n-1} e^{-x} \, dx = \Gamma(n). $$ Changing variables gives $$ \int_0^{\infty} x^{n-1} e^{-kx} \, dx = \frac{\Gamma(n)}{k^n}. $$

Euler now assumes that this still works if $k=p \pm iq$ is complex, provided $p>0$. (It does, but this needs an application of Cauchy's theorem.) We then have $$ \int_0^{\infty} x^{n-1} e^{-(p \pm iq)x} \, dx = \frac{\Gamma(n)}{(p \pm iq)^n}, $$ and if we write $p=f\cos{\theta}$, $q=f\sin{\theta}$, we can apply Euler's formula to obtain $$ \int_0^{\infty} x^{n-1} e^{-(p \pm iq)x} \, dx = \frac{\Gamma(n)}{f^n}(\cos{n\theta} \mp i\sin{n\theta}). $$ Adding and subtracting gives us the two integrals $$ \int_0^{\infty} x^{n-1} e^{-px} \cos{qx} \, dx = \frac{\Gamma(n)}{f^n}\cos{n\theta} \\ \int_0^{\infty} x^{n-1} e^{-px} \sin{qx} \, dx = \frac{\Gamma(n)}{f^n}\sin{n\theta} $$ The second is the one we care about: taking the limit as $n \to 0$, the left-hand side makes sense, and (here we deviate from Euler's infinitesimal discussion of $\Gamma$ and sine to avoid needless controversy) $ \Gamma(n)\sin{n\theta} \sim \frac{1}{n}n\theta = \theta $, so, converting back to $p$ and $q$, we find $$ \int_0^{\infty} \frac{1}{x} e^{-px} \sin{qx} \, dx = \theta = \arctan{\frac{q}{p}} $$

The result now follows almost as an afterthought, by putting $p=0$.

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The answer is correct.

A related technique. Recalling the Laplace transform

$$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$ We can use the following relation

$$ \begin{align} \int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt] L[f(t)] & = F(s) \\[6pt] L[g(t)] & = G(s)\end{align} $$


$$ G(u)=\frac{1}{u} \implies g(u)=1, $$


$$ f(u)= \sin(u) \implies F(u) = {\frac {1}{ \left( {u}^{2}+1 \right) }}. $$


$$ \int_0^\infty \frac{\sin u}{u} \, dx = \int_0^\infty \frac{1}{\left( {u}^{2}+1 \right)} \, du = \frac{\pi}{2}$$

Mhenni Benghorbal
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    @user1729: I don't think there is anything wrong with adding a 17th answer, so long as it contributes something new. On the surface, it appears that it does. Nevertheless, if one ponders where the LT relation above comes from, one will see that this method is precisely that of Aryahbata below, but packaged a little differently. If this method referenced Aryahbata's solution and pointed out how it could be formalized, and under what conditions, then this would be genuinely new. But as it doesn't, it appears to be a rehash. – Ron Gordon Aug 10 '13 at 00:30
  • @MhenniBenghorbal: correctness $\ne$ usefulness. The votes are based on usefulness. But you know this already. BTW the downvote is not mine. But if it were, how would that knowledge help you? – Ron Gordon Aug 10 '13 at 00:38
  • @RonGordon: I do not know why you people assume it is not useful? It is an approach to solve the problem and I used it before to solve other problems! It is totally correct and useful answer and simplifies the integral in a nice way as it was commented by one of the users beneath my related answer in the provided link. – Mhenni Benghorbal Aug 10 '13 at 04:24
  • @RonGordon I seems to me that this is just a fleshing out of Night Owl's answer (certainly, they have the same global idea). Perhaps I am mistaken? Also, for what it is worth, the downvote it not mine. – user1729 Aug 11 '13 at 15:09

I would like to present yet another simple proof that goes through Fourier series. However, we will need the following theorem; we denote by $S_n(x;f)$ the $n$-th partial sum of the Fourier series of $f(\in L^1[-\pi,\pi]$ and $2\pi$-periodic) at $x$. Then:

Theorem. Riemann's principle of localization. If $f\in L^1[-\pi,\pi]$, then $$ S_n(x,f)=\frac{1}{\pi}\int_{-\delta}^{\delta} f(x+t)\frac{\sin nt}{t}\, dt\, + o(1). \quad (\delta>0) $$ Now, if we pick the function $f(x)\equiv 1$, then $S_n(x,f)\equiv 1$ for all $n,x$. Thus, by Riemann's principle of localization: $$ 1=\frac{1}{\pi}\int_{-\delta}^{\delta}\frac{\sin nt}{t}\, dt \, + o(1) = \frac{2}{\pi}\int_0^{n\delta} \frac{\sin t}{t}\, dt + o(1). $$ Letting $n\to \infty$, we get $$ 1=\frac{2}{\pi}\int_0^{\infty}\frac{\sin t}{t}\, dt, $$ which yields the desired result.

Alberto Debernardi
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To express the integral in terms of the series expansion coefficients can be done by invoking Glaisher's theorem, which is a special case of Ramanujan's master theorem. If $f(x)$ is an even function such that $\int_0^{\infty}f(x) dx$ exists, and the series expansion $f(x) = \sum_{0}^{\infty}(-1)^n c_n x^{2n}$ is valid in a neighborhood of $x = 0$, then we have:

$$\int_0^{\infty} f(x)dx = \frac{\pi}{2}c_{-\frac{1}{2}}\tag{1}$$

where $c_{-\frac{1}{2}}$ is to be interpreted as an analytic continuation of $c_n$ obtained by replacing factorials by gamma functions. Glaisher derived his theorem in a non-rigorous way, later Ramanujan formulated his master theorem, which was later rigorously proven by Hardy. That rigorous proof then does involve the residue theorem, but the arguments by Ramanujan and the earlier arguments by Glaisher don't involve any complex analysis.

Also while superficially the formula for the integral looks analogous to what you can get from applying the residue theorem, i.e. that a real integral is proportional to an expansion coefficient, unlike the reside theorem there is now no contour in the complex plane to consider over which the integral has to vanish. So, even if there exist no contour for evaluating the integral using the residue theorem, as long as you got analytic expressions for the expansion coefficients, you'll be able to write down the integral.

It's easy to justify the formula on heuristic grounds, an argument similar to the original argument by Glaisher works as follows. One introduces the operator $E$ that acts on the expansion coefficients as:

$$E c_n = c_{n+1}\tag{2}$$

This then allows one to write $f(x)$ formally as:

$$f(x) = \sum_{n=0}^{\infty}(-1)^n c_n x^{2n} = \sum_{n=0}^{\infty}(-1)^n E^n c_0 x^{2n} = \frac{c_0}{1+E x^2}$$

Then if $E$ were a positive number, we would have:

$$\int_0^{\infty} \frac{c_0}{1+E x^2}dx = \frac{\pi}{2}E^{-\frac{1}{2}}c_0$$

One then assumes that this relation will still hold for $E$ the operator defined by Eq. (2), which implies the result given by Eq. (1).

For $f(x)=\dfrac{\sin(x)}{x}$, the $c_n$ are given by:

$$c_n = \frac{1}{(2n+1)!}$$

therefore $c_{-\frac{1}{2}} = 1$, the integral is therefore equal to $\dfrac{\pi}{2}$.

Count Iblis
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$$I=\int_0^\infty \frac{\sin x} x \, dx.$$

By the Schwinger parametrization we get

$$ I= \int_0^\infty \mathrm{d}t\int_0^\infty \sin{x}\exp(-t x)\, \mathrm{d}x.$$

The last integral can be evaluated by parts. Another simple way is using $\sin{x}=\Im\left[ \mathrm{e}^{-i x} \right]$:

$$\int_0^\infty \sin{x}\exp(-t x)\, \mathrm{d}x=\Im\int_0^\infty \mathrm{e}^{-(t-i)x}\, \mathrm{d}x=\frac{1}{1+t^2}.$$


$$I=\int_0^\infty \frac{\mathrm{d}t}{1+t^2}.$$

Here you can use again the Schwinger trick. However,


Shuch that

$$\int_0^\infty \frac{\sin x} x \, dx=\arctan(\infty)-\arctan(0)=\frac{\pi}{2}$$

Dinesh Shankar
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Another approach is to employ Laplace Transforms.

$$I = \int_{0}^{\infty}\frac{\sin(x)}{x}\, \mathrm dx.$$


$$I(t) = \int_{0}^{\infty}\frac{\sin(xt)}{x} \,\mathrm dx$$

Take the Laplace Transform to yield \begin{align*} \mathscr L[I(t)] &= \int_{0}^{\infty}\frac{\mathscr L[\sin(tx)]}{x}\,\mathrm dx\\ &= \int_{0}^{\infty}\frac{ 1}{x}\frac{x}{s^2 + x^2}\,\mathrm dx\\ &= \int_{0}^{\infty}\frac{1}{x^2 + s^2}\,\mathrm dx \\ &= \left[\frac{1}{s}\arctan\left(\frac{x}{s} \right) \right]_{0}^{\infty} \\ &= \frac{1}{s}\frac{\pi}{2} \end{align*}

And so, to solve $I(t)$ we take the inverse Laplace transform:

\begin{align*} I(t) &= \mathscr L^{-1}\left[\frac{1}{s}\frac{\pi}{2} \right] = \frac{\pi}{2}.1 = \frac{\pi}{2} \end{align*}


$$\int_{0}^{\infty} \frac{\sin(x)}{x}\mathrm dx = I(1) = \frac{\pi}{2}$$


Here's a "proof" I came up with when asked to use this integral for a physics homework problem in undergrad (I think we were expected to cite wolfram alpha). Perhaps worth adding for the novelty, if not the rigor, I think:

Define $$I(a) = \int_0^\infty \frac{\sin(2\pi ax)}{x}dx =\frac{i}{2} \int_{-\infty}^\infty \frac{e^{-2\pi iax}}{x} dx$$ for $a \in \mathbb{R}$. Then $$I'(a)=\pi \int_{-\infty}^\infty e^{-2\pi iax} dx =\pi \delta(a)$$ Where $\delta(x)$ is the Dirac delta "function". Integrating, $$I(a)=\pi \theta(a)+C$$ Where $\theta$ is the Heaviside function, for some $C \in \mathbb{R}$. Since $I(-a)=-I(a)$ from its definition, we must have for $a>0$ that $$\pi+C = I(a)=-I(-a)=-C $$ (since $\theta(a)=1$ and $\theta(-a)=0$). So $C=-\frac{\pi}{2}$, and our integral is $$I \left( \frac{1}{2 \pi} \right)=\pi-\frac{\pi}{2}=\frac{\pi}{2}$$ Perhaps someone knows a way to interpret this rigorously in terms of distributional derivatives and what have you, but I'm not there yet.

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Another "proof" using distributions and Fourier transforms.

We treat the integral as a sine transform, rewrite it as a Fourier transform and use the fact that the Fourier transform of $\frac{1}{x}$ is $-i\pi \, \operatorname{sign}(\xi)$: $$ \int_0^\infty \frac{\sin x}{x} dx = \frac{i}{2} \int_{-\infty}^{\infty} \frac{1}{x} e^{-ix} \, dx = \frac{i}{2} \left. \int_{-\infty}^{\infty} \frac{1}{x} e^{-i\xi x} \, dx \right|_{\xi=1} = \frac{i}{2} \left( -i\pi \, \operatorname{sign}(1) \right) = \frac{\pi}{2} . $$

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Define, for $\displaystyle x\geq 0,F(x)=\int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\cos(x\cos \theta)\,d\theta$


1)$\displaystyle F(x)=\frac{\pi}{2}-\int_0^x\frac{\sin \theta}{\theta}\,d\theta$

2)$\displaystyle \lim_{x\rightarrow\infty}F(x)=0$


  1. Observe that, \begin{align} F(0)&=\frac{\pi}{2}\\ F^\prime(x)&=-\int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\sin(\theta+x\cos \theta)\,d\theta\\ &=\left[\frac{\text{e}^{-x\sin \theta}\sin(x\cos \theta)}{x}\right]_{\theta=0}^{\theta=\frac{\pi}{2}}\\ &=-\frac{\sin x}{x} \end{align}

2)Observe that, for $\displaystyle 0\leq \theta\leq\frac{\pi}{2},0\leq \sin\theta<1$

For $x>0,0<a<1$, \begin{align}\left|F(x)\right|&\leq \int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\,d\theta\\ \int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\,d\theta&\overset{u=\sin\theta}=\int_0^1 \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du\\ &=\int_0^a \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du+\int_a^1 \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du\\ &\leq \int_0^a \frac{1}{\sqrt{1-u^2}}du+\text{e}^{-ax}\int_0^1 \frac{1}{\sqrt{1-u^2}}du\\ \end{align}


For all $0<a<1$, $\displaystyle \lim_{x\rightarrow +\infty}\int_0^1 \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du\leq \int_0^a \frac{1}{\sqrt{1-u^2}}du$ that is, $\displaystyle \lim_{x\rightarrow +\infty}\int_0^1 \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du=0$

(Thanks to Zephir for the help)

Thus, $\displaystyle \lim_{x\rightarrow \infty}F(x)=0$ and, $\boxed{\int_0^\infty\frac{\sin \theta}{\theta}\,d\theta=\frac{\pi}{2}}$

NB: I assume that $\displaystyle \int_0^\infty\frac{\sin \theta}{\theta}\,d\theta$ is converging (edit: no need to assume this)

Inspired by, Uebungsbuch zum Studium der höheren Analysis , a book by Oscar Schlömilch

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  • From $\int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\,d\theta\leq \frac{\pi}{2}\text{e}^{-x}+\frac{\pi}{2}x$, you cannot obtain $\lim_{x\to\infty}F(x)=0$. I think you have to use this inequality $\frac{2}{\pi}\theta\le\sin\theta\le\theta$. – xpaul Oct 21 '20 at 17:21
  • Of course, you're right. I have fixed, i hope, the mistake. – FDP Oct 22 '20 at 10:58
  • It's not yet fixed. Bad formula :( – FDP Oct 22 '20 at 11:01
  • I think it's fixed now. Thanks again for your concern. – FDP Oct 22 '20 at 15:16

By definition (Laplace Transform): \begin{equation*} F(s)=L\left[\frac{\sin(t)}{t}\right]=\int_{0}^{\infty}\frac{\sin(t)}{t}e^{-st}dt=\arctan^{-1}\left(\frac{1}{s}\right) \end{equation*} Then, for $s=0$, \begin{equation*} F(0)=\int_{0}^{\infty}\frac{\sin(t)}{t}dt = \lim_{s\to0}\arctan^{-1}\left(\frac{1}{s}\right)=\lim_{u\to\infty}\arctan^{-1}(u)=\frac{\pi}{2} \end{equation*}

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Here is an unusual approach using what physicists have recently called Integration by differentiation. Full justification for the results quoted below can be found in the following paper.

Define $\partial_y := \frac{d}{dy}$. The following integration by differentiation method has been introduced (there are others, see paper in the link for details) $$\int_0^\infty f(x) \, dx = \lim_{y \to 0^+} f(-\partial_y) \frac{1}{y}.\tag1$$ Something else we need is the following result. For a smooth function $f$ $$e^{a \partial_y} f(y) = f(y + a), \quad a \in \mathbb{C}. \tag2$$ Here $e^{a \partial_y}$ acts as a translation (shift) operator.

We are now in a position to proceed. Setting $f(x) = \sin x/x$ we have \begin{align*} \int_0^\infty \frac{\sin x}{x} dx &= \lim_{y \to 0^+} \frac{\sin (-\partial_y)}{-\partial_y} \frac{1}{y} \tag3\\ &= \lim_{y \to 0^+} \frac{\sin (\partial_y)}{\partial_y} \frac{1}{y}\\ &= \lim_{y \to 0^+} \frac{e^{i \partial_y} - e^{-i \partial_y}}{2i \partial_y} \frac{1}{y} \tag4\\ &= \frac{1}{2i} \lim_{y \to 0^+} \left (e^{i \partial_y} - e^{-i \partial_y} \right ) \log (y) \tag5\\ &= \frac{1}{2i} \lim_{y \to 0^+} \left (\log(y + i) - \log (y - i) \right ) \tag6\\ &= \frac{1}{2i} \lim_{y \to 0^+} \log \left (\frac{y + i}{y - i} \right ) \\ &= \lim_{y \to 0^+} \operatorname{arccot} (y) = \frac{\pi}{2}, \end{align*} as expected.


(3) Application of the integration by differentiation result given in (1).

(4) Expressed the sine function in terms of complex exponentials using: $\sin z = (e^{iz} - e^{-iz})/(2i)$,

(5) The antiderivative (the term $\frac{1}{\partial_y}$ is an antiderivative) of $\frac{1}{y}$ is $\log (y)$.

(6) Used the shift operator result given in (2).

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

By "closing" a contour in the first quadrant ( a quarter circle of radius $\ds{R}$ ):

\begin{align} \int_{0}^{R\ >\ 0}{\sin\pars{x} \over x}\,\dd x & = \Im\int_{0}^{R}{\expo{\ic x} - 1\over x}\,\dd x \\[5mm] & = -\,\Im\ \overbrace{\int_{0}^{\pi/2}{\exp\pars{\ic R\expo{\ic\theta}} - 1 \over R\expo{\ic\theta}}\,R\expo{\ic\theta}\ic\,\dd\theta} ^{\ds{\mbox{along the arc}}}\ -\ \Im\ \overbrace{\int_{R}^{0}{\expo{-y} - 1 \over \ic y}\,\ic\,\dd y} ^{\ds{\mbox{along the}\ y\ \mbox{axis}}} \\[5mm] & = -\,\Re\int_{0}^{\pi/2}\bracks{\exp\pars{\ic R\cos\pars{\theta}} \exp\pars{ -R\sin\pars{\theta}} - 1}\,\dd\theta \\[5mm] & = {\pi \over 2} - \Re\int_{0}^{\pi/2}\exp\pars{\ic R\cos\pars{\theta}} \exp\pars{ -R\sin\pars{\theta}}\,\dd\theta = \bbx{\pi \over 2} \end{align}

Note that

\begin{align} 0 & < \verts{\int_{0}^{\pi/2}\exp\pars{\ic R\cos\pars{\theta}} \exp\pars{ -R\sin\pars{\theta}}\,\dd\theta} < \int_{0}^{\pi/2}\exp\pars{-R\sin\pars{\theta}}\,\dd\theta \\[5mm] & < \int_{0}^{\pi/2}\exp\pars{-\,{2R \over \pi}\,\theta}\,\dd\theta = {\expo{-R} - 1 \over -2R/\pi}\,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\Large\to}\,\,\,{\large 0} \end{align}

Felix Marin
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$$ \begin{aligned} \text{Let }I(a)&=\int_{0}^{\infty} \frac{e^{-a x} \sin x}{x} d x \text {, where } a>0 \\ I^{\prime}(a)&=\int_{0}^{\infty} \frac{-x e^{-a x} \sin x}{x} d x \\ &=-\int_{0}^{\infty} e^{-a x} \sin x d x \\ &\stackrel{\text { IBP }}{=}-\frac{1}{a^{2}+1} \end{aligned} $$ Integrating w.r.t. $a$ yields $$ \begin{aligned} I(a) &=-\int \frac{d a}{a^{2}+1}+C=-\tan ^{-1} a+C \end{aligned} $$ Letting $a$ tends to infinity yields $C=\dfrac{\pi}{2}. $

$$ \therefore I(a)=-\tan ^{-1} a+\frac{\pi}{2} $$$$ \text {Hence } \boxed{ \int_{0}^{\infty} \frac{\sin x}{x} d x=I(0)=\frac{\pi}{2}} $$

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