Help find hard integrals that evaluate to $59$?

Question

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!

Now let's just hope that your father doesn't read math.stackexchange.com... — Tim Pietzcker, Jul 09 '12 at 06:34
Wiki-hammered as this is a bit list with a subjective (as hard as possible) answer. — Willie Wong, Jul 09 '12 at 08:33
Yeah... or your dad might know right away that the solution evaluates to 59... on his birthday... when he just turned 59. — phatfingers, Jul 10 '12 at 18:29
The fun is on proving it. So, it doesn't matter if the father sees this post or if he/she knows the answer is 59. — OR., Dec 19 '13 at 18:12

score 140 · Answer 1 · answered Jul 09 '12 at 17:41

140

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

answered Jul 09 '12 at 17:41

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. – 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 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. – 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.

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 : $$\frac 1{\cosh(t)}=\sum_{n=0}^\infty E_n\frac {t^n}{n!}$$ allows to rewrite the left part simply as $E_{2n}$. – Raymond Manzoni Jun 04 '13 at 08:56

score 90 · Accepted Answer · edited Jul 09 '12 at 15:07

90

You might try the following: $$ \frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx $$

edited Jul 09 '12 at 15:07

Community

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answered Jul 09 '12 at 04:43

Robert Israel

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Is that $$\ln(x^2)$$ or $$(\ln x)^2$$ ? – ypercubeᵀᴹ Jul 09 '12 at 12:06
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@ypercube: the second one. – J. M. ain't a mathematician Jul 09 '12 at 12:43
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Added to the card. Thanks very much! – Argon Jul 11 '12 at 01:11
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How does one start to solve this? – yiyi Dec 11 '12 at 08:50
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Does the indefinite integral of the integrand give a closed-form solution? – user85798 Oct 30 '13 at 08:29
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Using the Risch-Norman algorithm, the antiderivative is not elementary. Maple doesn't find a closed-form antiderivative. – Robert Israel Oct 30 '13 at 14:31

Alex Becker · Answer 3 · 2012-07-11T03:06:01.297

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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.

edited Jul 11 '12 at 03:06

answered Jul 09 '12 at 03:13

Alex Becker

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@PeterTamaroff Thanks. – Alex Becker Jul 09 '12 at 03:33
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Out of curiosity, why the downvote? – Alex Becker Jul 09 '12 at 06:29
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@JoachimSauer: It's probably due to my dumbness, but I don't see where the 59 is hidden in 108? 59*2=118, if you meant that? – Jakob S. Jul 09 '12 at 11:23
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@JoachimSauer That was actually a typo on my part. $108$ was meant to be $118$. – Alex Becker Jul 11 '12 at 03:07
@حكيمالفيلسوفالضائع It's only a conjecture? – MT_ May 02 '14 at 21:36
@MichaelT After some research I found that it isn't. – Hakim May 02 '14 at 21:41

score 24 · Answer 4 · answered Jul 09 '12 at 05:20

24

There's also

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

answered Jul 09 '12 at 05:20

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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

score 24 · Answer 5 · answered Jul 09 '12 at 07:41

24

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...

answered Jul 09 '12 at 07:41

J. M. ain't a mathematician

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Waiting on your edit as promised.... :-) – msh210 Jun 13 '13 at 05:16
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Ah, shoot; let me look for my notes on this... if memory serves, I trolled through OEIS and looked for generating functions. – J. M. ain't a mathematician Jun 13 '13 at 05:24
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http://i.imgur.com/8PfsQ.png – wchargin May 12 '14 at 00:09
Almost 8 years later. You still haven't posted the solution. Please... – Laxmi Narayan Bhandari Oct 10 '21 at 02:15
Still waiting for that solution. – libanobovich Jun 05 '22 at 17:52
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Still waiting for the solution! :D – Max0815 Jan 22 '23 at 02:01

Help find hard integrals that evaluate to $59$?

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