Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form? $$\int_0^\infty\frac{\cos x}{1+x^2}\,\mathrm{d}x$$

Michael Hardy
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Martin Gales
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    Complex analysis methods: http://math.stackexchange.com/questions/100616/how-to-evaluate-this-integral – Aryabhata Jan 20 '12 at 02:52

12 Answers12


This can be done by the useful technique of differentiating under the integral sign.

In fact, this is exercise 10.23 in the second edition of "Mathematical Analysis" by Tom Apostol.

Here is the brief sketch (as laid out in the exercise itself).

Let $$ F(y) = \int\limits_{0}^{\infty} \frac{\sin xy}{x(1+x^2)} \ dx \ \ \text{for} \quad\quad y > 0$$

Show that

$\displaystyle F''(y) - F(y) + \pi/2 = 0$ and hence deduce that $\displaystyle F(y) = \frac{\pi(1-e^{-y})}{2}$.

Use this to deduce that for $y > 0$ and $a > 0$

$$\displaystyle \int_{0}^{\infty} \frac{\sin xy}{x(x^2 + a^2)} \ dx = \frac{\pi(1-e^{-ay})}{2a^2}$$


$$\int_{0}^{\infty} \frac{\cos xy}{x^2 + a^2} dx = \frac{\pi e^{-ay}}{2a}$$

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Since $$\frac{x}{1+x^2}=\int_0^\infty e^{-y}\sin (xy) \, dy,$$ we have that $$I=\int_0^\infty \frac{\cos bx}{1+x^2} \, dx=\int_0^\infty \frac{\cos bx}{x} \, dx \int_0^\infty e^{-y}\sin (xy)\, dy.$$ Changing the order of integration (which can be justified by the standard method) yields $$I=\int_{0}^{\infty}e^{-y} \, dy \int_0^\infty \frac{\sin xy}{x} \cos bx \, dx.$$ The calculation of the integral (a.k.a. the discontinuous Dirichlet factor) $$\int_0^\infty \frac{\sin xy}{x} \cos bx \, dx = \begin{cases}0, & 0 < y < b \\\ \ \\\ \pi/2, & 0 < b < y, \end{cases}$$ can be easily reduced to the calculation of the standard Dirichlet integral. Therefore, $$I=\frac{\pi}{2}\int_{b}^{\infty}e^{-y} \, dy = \frac{\pi}{2}e^{-b}.$$

Michael Hardy
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Andrey Rekalo
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  • Very inspiring solution! I learned a lot. Thanks! – Martin Gales Nov 09 '10 at 06:10
  • . . . "which can be justified by the standard method" - it can be justified here, but not as routinely as you seem to suggest: the double integral does not quite converge absolutely, so Fubini is not automatic. – Noam D. Elkies Jun 24 '16 at 06:55
  • And how does one evaluate the Dirichlet integral without complex analysis? Feynman's Trick is one way forward. But why not simply use Feynman's Trick to begin and save yourself the unnecessary work herein? – Mark Viola Aug 22 '16 at 17:02

These are the methods I use to evaluate $$ \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx $$ and post it on Brilliant.org as a solution of similar problem. You can use the similar technique to evaluate $$ \int_0^{\infty}\frac{\cos x}{x^2+1}\,dx. $$

Method 1:

Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by $$ \begin{align} F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{0}^{t=v}\\ &=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\ &=\frac{2a}{\omega^2+a^2}. \end{align} $$ Next, the inverse Fourier transform of $F(\omega)$ is $$ \begin{align} f(t)=\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\\ e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{\omega^2+a^2}e^{i\omega t}\,d\omega\\ \frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{\omega^2+a^2}\,d\omega.\tag1 \end{align} $$ Now, rewrite $$ \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\mathbb{Re}\left(e^{2ix}\right)}{x^2+2^2}\,dx.\tag2 $$ Comparing $(2)$ to $(1)$ yield $t=2$ and $a=2$. Thus, $$ \begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx &=\frac{1}{2}\frac{\pi e^{-2\cdot|2|}}{2}\\ &=\frac{\pi}{4e^4}\\ \end{align} $$ and $$ \Large\color{blue}{\int_0^{\infty}\frac{\cos x}{x^2+1}\,dx=\frac{\pi}{2e}}. $$

Method 2:

Note that: $$ \int_{y=0}^\infty e^{-(x^2+4)y}\,dy=\frac{1}{x^2+4}, $$ therefore $$ \int_{x=0}^\infty\int_{y=0}^\infty e^{-(x^2+4)y}\cos2x\,dy\,dx=\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx $$ Rewrite $\cos2x=\Re\left(e^{-2ix}\right)$, then $$ \begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx&=\int_{x=0}^\infty\int_{y=0}^\infty e^{-(x^2+4)y}\cos2x\,dy\,dx\\ &=\int_{y=0}^\infty\int_{x=0}^\infty e^{-(yx^2+2ix+4y)}\,dx\,dy\\ &=\int_{y=0}^\infty e^{-4y} \int_{x=0}^\infty e^{-(yx^2+2ix)}\,dx\,dy. \end{align} $$ In general $$ \begin{align} \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\int_{x=0}^\infty \exp\left(-a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ \end{align} $$ Let $u=x+\frac{b}{2a}\;\rightarrow\;du=dx$, then $$ \begin{align} \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\ \end{align} $$ The last form integral is Gaussian integral that equals to $\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}$. Hence $$ \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right). $$ Thus $$ \int_{x=0}^\infty e^{-(yx^2+2ix)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(\frac{(2i)^2}{4y}\right)=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(-\frac{1}{y}\right). $$ Next $$ \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy. $$ In general $$ \begin{align} \int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dv\\ &=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ \end{align} $$ The trick to solve the last integral is by setting $$ I=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv. $$ Let $t=-\frac{1}{v}\sqrt{\frac{b}{a}}\;\rightarrow\;v=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dv=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then $$ I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt. $$ Let $t=v\;\rightarrow\;dt=dv$, then $$ I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt. $$ Adding the two $I_t$s yields $$ 2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt. $$ Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then $$ I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}. $$ Thus $$ \begin{align} \int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ &=\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}\\ \end{align} $$ and $$ \begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx&=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy\\ &=\frac{\sqrt{\pi}}{2}\cdot\sqrt{\frac{\pi}{4}}e^{-2\sqrt{4\cdot1}}\\ &=\frac{\pi}{4e^4}. \end{align} $$ Hence $$ \Large\color{blue}{\int_0^{\infty}\frac{\cos x}{x^2+1}\,dx=\frac{\pi}{2e}}. $$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\cos\pars{x} \over 1 + x^{2}}\dd x:\ {\large ?}}$ \begin{align} \mbox{Lets}\quad\fermi\pars{\mu} & \equiv \half\int_{-\infty}^{\infty}{\cos\pars{\mu x} \over 1 + x^{2}}\,\dd x \\[5mm] \quad\mbox{such that}\quad & \left\lbrace\begin{array}{rcl} \ds{\int_{0}^{\infty}{\cos\pars{x} \over 1 + x^{2}}\dd x} & \ds{=} & \ds{\fermi\pars{1}} \\[2mm] \ds{\fermi\pars{0}} & = & \ds{\pi \over 2} \end{array}\right. \end{align}

\begin{align} \fermi''\pars{\mu} & = \half\int_{-\infty}^{\infty} {-x^{2}\cos\pars{\mu x} \over 1 + x^{2}}\,\dd x \\[5mm] & = -\pi\,\Re\int_{-\infty}^{\infty}\expo{\ic\mu x} \,{\dd x \over 2\pi} +\fermi\pars{\mu} \\[5mm]& \implies\quad\fermi''\pars{\mu} - \fermi\pars{\mu} = -\pi\,\delta\pars{\mu} \end{align}

The differential equation is equivalent to: $$\left\lbrace \begin{array}{rcl} \fermi''\pars{\mu} - \fermi\pars{\mu} = 0 & \mbox{if} & \mu \not= 0 \\[2mm] \fermi'\pars{0^{+}} - \fermi'\pars{0^{-}} & = & -\pi \end{array}\right. $$ When $\ds{\mu \not= 0}$, the solutions are linear combinations of $\ds{\expo{\pm\mu}}$. Since $\ds{\fermi\pars{0} = {\pi \over 2}}$and the solution is continuos at $\ds{\mu = 0}$ and finite, we'll get: $$ \fermi\pars{\mu} = {\pi \over 2}\,\expo{-\verts{\mu}} $$ It satisfies $\ds{\fermi'\pars{0^{+}} - \fermi'\pars{0^{-}} = \pars{-\,{\pi \over 2}} - \pars{{\pi \over 2}} = -\pi}$

$$\color{#44f}{\large \int_{0}^{\infty}{\cos\pars{x} \over 1 + x^{2}}\dd x} =\fermi\pars{1} = {\pi \over 2}\,\expo{-\verts{1}}= \color{#44f}{\large{\pi \over 2\expo{}}} $$

Felix Marin
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The method of brackets, which is an extension of Ramanujan's master theorem, can be used to evaluate this classic integral.

The term bracket refers to the assignment of the symbol $\langle a \rangle$ to the divergent integral $\int_{0}^{\infty} x^{a -1} \, \mathrm{d}x$.

You can read about the method in the following papers:

Definite integrals by the method of brackets. Part 1

The method of brackets. Part 2: Examples and applications

Evaluation of entries in Gradshteyn and Ryzhik employing the method of brackets

On the Method of Brackets: Rules, Examples, Interpretations and Modifications

The hypergeometric representation of the cosine function is $$ \, _0F_{1} \left(; \frac{1}{2}; - \frac{x^{2}}{4} \right) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \, \frac{\Gamma \left( \frac{1}{2} \right)}{\Gamma \left(n+\frac{1}{2} \right)} \left(\frac{x^{2}}{4} \right)^{n} = \sum_{n=0}^{\infty} \phi_{n} \, \frac{\Gamma \left( \frac{1}{2} \right)}{\Gamma \left(n+\frac{1}{2} \right)} \left(\frac{x^{2}}{4} \right)^{n}. $$

(This representation can be derived from the Maclaurin series of $\cos (x)$ by using the duplication formula for the gamma function).

And according to Rule 3.1 on page 8 of the first paper, the function $ \frac{1}{1+x^{2}}$ is assigned the bracket series $$\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \phi_{k,m} \, x^{2m} \langle k+m+1 \rangle. $$

Therefore, the integral $ \int_{0}^{\infty} \frac{\cos x}{1+x^{2}} \, \mathrm{d}x$ corresponds with the bracket series $$ \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \phi_{k,m,n} \, \frac{\Gamma \left(\frac{1}{2} \right)}{\Gamma \left(n+\frac{1}{2}\right)} \frac{1}{4^{n}} \langle k+m+1\rangle \langle 2m+2n+1 \rangle. \tag{1}$$

To evaluate $(1)$, first let $k$ be a free parameter.

The brackets then vanish if $m=-k-1$ and $n= k + \frac{1}{2}$.

So according to Rule 3.3 in the first paper, the contribution to the integral is

$$\begin{align}\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \frac{\Gamma \left(\frac{1}{2} \right)}{\Gamma (k+1 )} \, \frac{\Gamma (k+1) \Gamma \left(-k-\frac{1}{2} \right)}{4^{k+ 1/2}} &= \frac{\sqrt{\pi}}{4} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \frac{1}{4^{k}} \frac{\pi (-1)^{k-1} }{\Gamma \left(k + \frac{3}{2} \right)} \\ &= - \frac{\pi \sqrt{\pi} }{4} \sum_{k=0}^{\infty} \frac{1}{k!} \frac{1}{4^{k}} \frac{2^{2(k+1)-1} \Gamma(k+1)}{ \sqrt{\pi} \, \Gamma(2k+2)} \\ &= -\frac{\pi}{2} \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} \\ &= -\frac{\pi}{2} \, \sinh (1). \end{align}$$

(I used the reflection formula for the gamma function on the first line, and then I used the duplication formula for the gamma function on the second line.)

Now let $m$ be a free parameter.

The brackets then vanish if $k = -m-1$ and $n=-m -\frac{1}{2}$.

So according to Rule 3.3, the contribution to the integral is $$ \frac{1}{2} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!} \frac{\Gamma \left(\frac{1}{2} \right)}{{\color{red}{\Gamma (-m)}}} \frac{\Gamma(m+1) \Gamma \left(m+ \frac{1}{2} \right)}{4^{-m-1/2}} =0.$$

Finally let $n$ be a free parameter.

The brackets then vanish if $k = - n - \frac{1}{2}$and $m= n - \frac{1}{2}$.

So according to Rule 3.3, the contribution to the integral is

$$\begin{align}\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{\Gamma \left(\frac{1}{2} \right)}{\Gamma (n + \frac{1}{2} )} \, \frac{\Gamma \left(n + \frac{1}{2} \right) \Gamma \left(\frac{1}{2}-n \right) }{4^{n}} &= \frac{\sqrt{\pi}}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{1}{4^{n}} \frac{\pi (-1)^{n}}{\Gamma \left(n+ \frac{1}{2} \right)} \\ &= \frac{\pi \sqrt{\pi}}{2} \sum_{n=0}^{\infty} \frac{1}{n!} \frac{1}{4^{n}} \frac{2^{2(n+1/2)-1} \Gamma(n+1)}{\sqrt{\pi} \, \Gamma(2n+1)} \\ &= \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{(2n)!} \\ &= \frac{\pi}{2} \, \cosh (1). \end{align}$$

Rule 3.4 in the first paper says that we should add these three contributions to get the value of the integral.

Therefore, $$\int_{0}^{\infty} \frac{\cos(x)}{1+x^{2}} \, \mathrm{d}x = - \frac{\pi}{2} \sinh(1) + 0 + \frac{\pi}{2} \cosh(1) = \frac{\pi}{2e}.$$

Random Variable
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Another way forward is to use Feynman's Trick. To that end, let $I(a)$ be the integral

$$I(a,b)=\int_0^\infty e^{-ax}\frac{\cos(bx)}{1+x^2}\,dx \tag 1$$

for $a\ge 0$. Then, for $a>0$ we have

$$\begin{align} \frac{\partial^2 I(a,b)}{\partial b^2}&=-\int_0^\infty e^{-ax}\frac{x^2\cos(bx)}{1+x^2}\,dx\\\\ &=I(a,b)-\frac{a}{a^2+b^2} \end{align}$$

Therefore, using $\lim_{b\to \infty}I(a,b)=0$ (apply the Riemann-Lebesgue Lemma), we find that for $a>0$

$$I(a,b)=C(a)e^{-|b|}+\frac12\int_0^\infty e^{-|b-x|}\frac{a}{a^2+x^2}\,dx \tag 2$$

for some function $C(a)$.

When $b=0$ and $a\to 0^+$, $I(a,b)$ as given by $(2)$ is $I(0^+,0)=C(0^+)+\frac{\pi }{4}$, where $I(0^+,0)$ as given by $(1)$ is $I(0^+,0)=\frac{\pi}{2}$. Therefore, we find that $C(0^+)=\frac{\pi}{4}$.

Finally, when $b=1$ we have from $(1)$,

$$\lim_{a\to 0^+}I(a,1)=I(0,1) \tag 3$$

whereas we have from $(2)$, we have

$$\lim_{a\to 0^+}I(a,1)=\frac{\pi}{2e}\tag 4$$

Comparing $(3)$ and $(4)$ yields


as expected!

Mark Viola
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The previous answer is not correct. If you use the Taylor expansion of cosine and integrate termwise you consider integrals of the following form: \begin{eqnarray} \int_{0}^{\infty} \frac{x^{a} \ dx}{1 + x^{2}} = \tfrac{\pi}{2} \sec (\tfrac{\pi a}{2}) \end{eqnarray} which is only well-defined if $-1 < a < 1$.

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Let $\lambda\in\mathbb{R}$, set $$I(\lambda)=\int_{-\infty}^{\infty}{\cos(\lambda x)\over x^2+1}dx$$ we use integrate by parts, writing
$$u=\frac{1}{{{x}^{2}}+1}\quad,\quad dv=\cos (\lambda x)$$ we have $$I(\lambda )=\frac{\sin (\lambda x)}{\lambda ({{x}^{2}}+1)}\left| \begin{matrix} \infty \\ -\infty \\ \end{matrix} \right.+\frac{2}{\lambda }\int_{-\infty }^{+\infty }{\frac{\sin (\lambda x)}{{{({{x}^{2}}+1)}^{2}}}}\,dx $$ as a result $$\lambda I(\lambda )=2\int_{-\infty }^{\infty }{\frac{x\sin \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx \,.\quad(1)$$ By differentiate with respect $\lambda$ to get $$\lambda \frac{dI}{d\lambda }+I(\lambda )=2\int_{-\infty }^{\infty }{\frac{{{x}^{2}}\cos \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx=\underbrace{2\int_{-\infty }^{\infty }{\frac{\cos \lambda x}{{{x}^{2}}+1}\,}dx}_{2I(\lambda )}-2\int_{-\infty }^{\infty }{\frac{\cos \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx$$ therefore $$\lambda \frac{dI}{d\lambda }-I(\lambda )=-2\int_{-\infty }^{\infty }{\frac{\cos \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx$$
and $$\lambda \frac{{{d}^{2}}I}{d{{\lambda }^{2}}}=2\int_{-\infty }^{\infty }{\frac{x\sin \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx.\quad(2)$$ $(1)$ and $(2)$ $$\frac{{{d}^{2}}I(\lambda)}{d{{\lambda }^{2}}}- I(\lambda )=0$$ thus $$I(\lambda)=c_1e^{\lambda}+c_2e^{-\lambda}$$ on the other hand \begin{align} & I(0)={{c}_{1}}+{{c}_{2}}=\int_{-\infty }^{+\infty }{\frac{1}{{{x}^{2}}+1}}\,dx=\pi \,\,\,\,\Rightarrow \,\,{{c}_{1}}+{{c}_{2}}=\pi \, \\ & I(\lambda )=\frac{2}{\lambda }\int_{-\infty }^{+\infty }{\frac{x\sin \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,\,}dx\,\,\,\Rightarrow \,\,\underset{\lambda \to \infty }{\mathop{\lim }}\,I(\lambda )=0\,\,\,\Rightarrow \,{{c}_{1}}=0 \\ \end{align} then $$I(\lambda )=\pi {{e}^{-\lambda }}$$ set $\lambda=1$, we have $$I(1)=\int_{-\infty}^{\infty}{\cos( x)\over x^2+1}dx=\frac{\pi}{e}$$ so $$\int_{0}^{\infty}{\cos( x)\over x^2+1}dx=\frac{\pi}{2e}$$

Behrouz Maleki
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Glaisher's theorem, which is a special case of Ramanujan's master theorem, can also be used to compute this integral. If a function has a series expansion of the form:

$$f(x) = \sum_{k=0}^{\infty}(-1)^k c_{k}x^{2k}$$


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

Here $c_{-\frac{1}{2}}$ is obtained by analytically continuing $c_k$, e.g. by writing $c_k$ in terms of gamma functions (the rigorous way to do this is to apply the rigorous version of Ramanujan's master theorem). The above expression is only valid if the integral converges. For $f(x)=\frac{\cos(x)}{1+x^2}$, we have:

$$c_k = \sum_{j=0}^{k}\frac{1}{(2j)!}$$

We can obtain the analytic continuation of $c_k$ by considering the limiting value at infinity:

$$\lim_{k\to\infty}c_{k} = \cosh(1)$$

and by imposing the recursion relation

$$c_{k+1} = c_{k} + \frac{1}{(2k+2)!}$$

for general $k$. This means that for general $k$ we have:

$$c_k + \sum_{j=k+1}^{\infty}\frac{1}{(2j)!} = \cosh(1)$$

To compute $c_{-\frac{1}{2}}$, we thus need to evaluate the summation:

$$\sum_{j=\frac{1}{2}}^{\infty}\frac{1}{(2j)!} = \sum_{j=0}^{\infty}\frac{1}{(2(j+\frac{1}{2}))!} = \sum_{j=0}^{\infty}\frac{1}{(2j+1)!} = \sinh(1)$$


$$c_{-\frac{1}{2}} = \exp(-1)$$

and the integral is given by $\frac{\pi}{2} \exp(-1)$.

Count Iblis
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Recall that, if we consider the Fourier transform $$\mathcal Ff (a) =\int_\Bbb R e^{-ia x}f(x)dx$$ then its Fourier inverse is defined as $$\mathcal F^{-1}f (x) =\frac{1}{2\pi}\int_\Bbb R e^{it x}f(t)dt.$$

But we have, \begin{split} \mathcal F(e^{-|t|})(x) = \int_{-\infty}^{\infty}e^{-|t|}e^{-ix t}\,dt &=&\int_{-\infty}^{0}e^{t}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-t}e^{-ix t}\,dt\\ &=&\left[ \frac{e^{(1-ix)t}}{1-ix} \right]_{-\infty}^0-\left[\frac{e^{-(1+ix)t}}{1+ix} \right]_{0}^{\infty}\\ &=&\frac{1}{1-ix}+\frac{1}{1+ix}\\ &=&\frac{2}{x^2+1}. \end{split} Then, $$ \begin{align} e^{-|a|}=\mathcal F^{-1}\left( \frac{2}{x^2+1}\right)(a) &=\frac{1}{2\pi}\int_\Bbb R \frac{2}{x^2+1}e^{ix a}\,dx = \frac{1}{\pi}\int_\Bbb R\frac{e^{ix a}}{x^2+1}\,dx \\&=\frac{1}{\pi}\int_\Bbb R\frac{\cos a x}{x^2+1}\,dx = \frac{2}{\pi}\int_0^\infty\frac{\cos ax}{x^2+1}\,dx \end{align} $$ Given that, as $x\mapsto\sin ax $ is an old function we have, $$\int_\Bbb R \frac{\sin{a x}}{x^2+1}dx= 0.$$

Thus we have, $$ \int_0^\infty\frac{\cos ax}{x^2+1}\,dx =\frac{\pi}{2}e^{-|a|} $$

Guy Fsone
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I am surprised nobody has written about Schwinger's method, which can be used to evaluate loop integrals in quantum field theory, and which nicely computes this integral in combination with Glasser's master theorem. First notice that the integral can be re-written as $$\int_0^\infty \cos(x)\int_0^\infty e^{-u(1+x^2)}\,du\,dx.$$ Now, switch the order of integration and extend the range of one of the integral to the entire real line, gaining a factor of $1/2$: $$\frac{1}{2}\int_0^\infty e^{-u}\int_{-\infty}^\infty e^{-x^2+ix}\,dx\,du.$$ Complete the square in the exponential, and evaluate the now Gaussian integral (use the analyticity of $e^{-z^2}$ to bring the line of integration back to the real line) to obtain $$\sqrt{\pi}\int_0^\infty \frac{du}{2\sqrt{u}}\,\exp-\left(u-\frac{1}{4u}\right).$$ The time is ripe for a substitution $x=\sqrt{u}$, bringing us to $$\frac{\sqrt{\pi}}{2}e^{-1}\int_{-\infty}^\infty dx\, \exp-\left(x-\frac{1}{2x}\right)^2.$$ Glasser's master theorem now tells us how to compute this last integral, reducing it to a Gaussian, and the original integral is just $$\frac{\pi}{2}e^{-1}.$$

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$$\int_{0}^{\infty}e^{-t}\sin (xt)dt=\frac{1}{1+x^2},$$

And $$\int_{0}^{\infty}\frac{\cos{mx}}{1+x^2}dx=\int_{0}^{\infty}dx\int_{0}^{\infty}dt\text{ }\big(e^{-t}\cos{(mx)}\cos{(xt)}\big),$$

$$=\int_{0}^{\infty}dx\int_{0}^{\infty}dt\text{ }\big(e^{-t}\frac{\cos{(m+t)x}+\cos(m-t)x}{2} \big)$$

where $\int_{0}^{\infty}\cos{Qa da=\pi\delta(Q)}$, and $\delta$ is Dirac-Delta function.

$$=\int_{0}^{\infty}dt\text{ }\frac{e^{-t}}{2}\big(\pi\delta(m+t)+\pi\delta(m-t) \big)$$

hence: $$\int_{0}^{\infty}\frac{\cos{mx}}{1+x^2}dx=0+\frac{\pi}{2}e^{-|m|},$$

plug in $m=1$:


Reference: Feynman's Lectures on Mathematical Methods (lecture notes) Lec 3,4 and 5.

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