# Help find hard integrals that evaluate to $59$?

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!

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Now let's just hope that your father doesn't read math.stackexchange.com... – Tim Pietzcker Jul 9 '12 at 6:34
Wiki-hammered as this is a bit list with a subjective (as hard as possible) answer. – Willie Wong Jul 9 '12 at 8:33
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. – Mlazhinka Shung Gronzalez LeWy Dec 19 '13 at 18:12

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

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Is that $$\ln(x^2)$$ or $$(\ln x)^2$$ ? – ypercube Jul 9 '12 at 12:06
@ypercube: the second one. – J. M. Jul 9 '12 at 12:43
Added to the card. Thanks very much! – Argon Jul 11 '12 at 1:11
How does one start to solve this? – yiyi Dec 11 '12 at 8:50
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

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

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With added benefit of having very few arbitrary-looking constants. – Alex Feinman Jul 9 '12 at 18:27
@Alex: should be even better in 2 years! :-) – Raymond Manzoni Jul 9 '12 at 18:37
I like this Easy to read, no crazy exponents yet nontrivial solution. – Chad Jul 9 '12 at 19:12
Wolfram Alpha timed out trying to evaluate it. – Joe Z. Mar 14 '13 at 14:57
@Raymond 59 = 61 - 2? – Joe Z. Mar 15 '13 at 12:29

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.

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@PeterTamaroff Thanks. – Alex Becker Jul 9 '12 at 3:33
Out of curiosity, why the downvote? – Alex Becker Jul 9 '12 at 6:29
@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 9 '12 at 11:23
@JoachimSauer That was actually a typo on my part. $108$ was meant to be $118$. – Alex Becker Jul 11 '12 at 3:07
@حكيمالفيلسوفالضائع It's only a conjecture? – Soke May 2 '14 at 21:36

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...

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Waiting on your edit as promised.... :-) – msh210 Jun 13 '13 at 5:16
Ah, shoot; let me look for my notes on this... if memory serves, I trolled through OEIS and looked for generating functions. – J. M. Jun 13 '13 at 5:24
i.imgur.com/8PfsQ.png – WChargin May 12 '14 at 0:09

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 9 '12 at 21:13
@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. – user12014 Jul 9 '12 at 21:17