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My father and I, on birthday cards, give mathematical equations for each others new age. This year, my father will be turning $59$.

I want to try and make a definite integral that equals $59$. So far I can only think of ones that are easy to evaluate. I was wondering if anyone had a definite integral (preferably with no elementary antiderivative) that is difficult to evaluate and equals $59$? Make it as hard as possible, feel free to add whatever you want to it!

Cooperation
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Argon
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5 Answers5

135

compact : $$\int_0^\infty \frac{(x^4-2)x^2}{\cosh(x\frac{\pi}2)}\,dx$$

Raymond Manzoni
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    With added benefit of having very few arbitrary-looking constants. – Alex Feinman Jul 09 '12 at 18:27
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    @Alex: should be even better in 2 years! :-) – Raymond Manzoni Jul 09 '12 at 18:37
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    I like this Easy to read, no crazy exponents yet nontrivial solution. – Chad Jul 09 '12 at 19:12
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    Wolfram Alpha timed out trying to evaluate it. – Joe Z. Mar 14 '13 at 14:57
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    @Joe: hint for fast evaluation : use the generating function for [Euler numbers](http://en.wikipedia.org/wiki/Euler_number). – Raymond Manzoni Mar 15 '13 at 12:18
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    @Raymond 59 = 61 - 2? – Joe Z. Mar 15 '13 at 12:29
  • @Joe: yes easy isn't it? :-) (and that's why it will be simpler in 2014!) – Raymond Manzoni Mar 15 '13 at 12:30
  • @RaymondManzoni There should be a lemma I am missing, but why $\int t^{2n}/cosh(t \pi/2) = |E_{2n}|$ ? – user10676 Jun 03 '13 at 14:44
  • A derivation without using the generating function : \begin{align} \frac {I(m)}2:=\int_0^\infty \frac{t^m}{2\,\cosh(t)}dt&=\int_0^\infty \frac{t^m\;e^{-t}}{1+e^{-2t}}dt\\ &=\int_0^\infty \sum_{k=0}^\infty (-1)^k\;t^m\;e^{-(2k+1)t}\;dt\\ &=\sum_{k=0}^\infty\frac {(-1)^k}{(2k+1)^m}\int_0^\infty u^m\;e^{-u}\frac{du}{2k+1}\;\\ &=\Gamma(m+1)\sum_{k=0}^\infty\frac {(-1)^k}{(2k+1)^{m+1}}\\ &=\Gamma(m+1)\,\beta(m+1)\\ \end{align} And the properties of the [Dirichlet Beta function](http://en.wikipedia.org/wiki/Dirichlet_beta_function) applied to $m=2n\,$ give : – Raymond Manzoni Jun 04 '13 at 01:56
  • $$I(2n)=(2n)!\;\beta(2n+1)=(-1)^n\;E_{2n}\;\left(\frac {\pi}2\right)^{2n+1}$$ while the indefinite integral returned by Alpha for $n=3$ looks like [this](http://www.wolframalpha.com/input/?i=int%28x%5E6/cosh%28x%29,x%29). – Raymond Manzoni Jun 04 '13 at 01:58
  • @user10676: There is indeed a lemma : $$\int_0^\infty \frac {\cosh(a\,x)}{\cosh(b\,x)}\;dx=\frac {\pi}{2b}\sec\left(\frac{\pi\,a}{2\,b}\right)$$ that we will use written as (with $b:=\pi/2$) : $$\frac 1{\cosh(t)}=\int_0^\infty \frac {\cos(t\,x)}{\cosh(\pi\,x/2)}\;dx$$ These formulas may be proved with [this method](http://math.stackexchange.com/questions/171073/how-to-evaluate-these-integrals-by-hand/171088#171088). This implies : $$\left(\frac d{dt}\right)^{2n}\frac 1{\cosh(t)}=(-1)^n\int_0^\infty \frac {x^{2n}\cos(t\,x)}{\cosh(\pi\,x/2)}\;dx$$ Conclude with the limit $\lim_{t\to 0}$. – Raymond Manzoni Jun 04 '13 at 08:17
  • since the [Euler numbers generating function](http://en.wikipedia.org/wiki/Euler_number) : $$\frac 1{\cosh(t)}=\sum_{n=0}^\infty E_n\frac {t^n}{n!}$$ allows to rewrite the left part simply as $E_{2n}$. – Raymond Manzoni Jun 04 '13 at 08:56
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You might try the following: $$ \frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx $$

Robert Israel
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Combining an very difficult infinite sum with the indefinite integral of $\sin(x)/x$ over $\mathbb R$, which has no elementary antiderivative, gives

$$\frac{118\sqrt{2}}{9801}\int_{\mathbb R} \left(\sum_{k=0}^\infty \left(\frac{(4k)!(1103+26390k)}{(k!)^4396^{4k}}\frac{\sin x}{x}\right)\right)dx=59\cdot \frac{1}{\pi}\cdot \pi=59$$

which should be tough enough to stump anyone who hasn't seen them before.

Alex Becker
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Somewhat complicated, but...

$$\begin{align*}\frac{12}{\pi}\int_0^{2\pi} \frac{e^{\frac12\cos\,t}}{5-4\cos\,t}&\left(2\cos \left(t-\frac{\sin\,t}{2}\right)+3\cos\left(2t-\frac{\sin\,t}{2}\right)+\right.\\&\left.14\cos\left(3t-\frac{\sin\,t}{2}\right)-8\cos\left(4t-\frac{\sin\,t}{2}\right)\right)\mathrm dt=59\end{align*}$$

As a hint on how I obtained this integral, I used Cauchy's differentiation formula on a certain function (I'll edit this answer later to reveal that function), and took the real part...

J. M. ain't a mathematician
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There's also

$$\int_0^\infty \! x^3 e^{-(118)^{-1/2}x^2} \, dx$$

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    The 118 is a bit too obvious, assuming it's not just coincidence that 118/2 = 59 – Random832 Jul 09 '12 at 21:13
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    @Random832 True, but I figured I'd write down an integral that gives his father a reasonable chance of figuring it out... ...at least I myself would struggle with some of the other alternatives. –  Jul 09 '12 at 21:17
  • Yeah! Our objective is not to teach him mathematics for that instance rather to provide him with the pleasure, challenge and thrill of solving such integral!!! Isn't it.... – Cooperation Jan 12 '21 at 04:43