# 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!

• Now let's just hope that your father doesn't read math.stackexchange.com... Jul 9, 2012 at 6:34
• Wiki-hammered as this is a bit list with a subjective (as hard as possible) answer. Jul 9, 2012 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. Jul 10, 2012 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, 2013 at 18:12

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

• With added benefit of having very few arbitrary-looking constants. Jul 9, 2012 at 18:27
• @Alex: should be even better in 2 years! :-) Jul 9, 2012 at 18:37
• I like this Easy to read, no crazy exponents yet nontrivial solution.
Jul 9, 2012 at 19:12
• Wolfram Alpha timed out trying to evaluate it. Mar 14, 2013 at 14:57
• @Raymond 59 = 61 - 2? Mar 15, 2013 at 12:29

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

• Is that $$\ln(x^2)$$ or $$(\ln x)^2$$ ? Jul 9, 2012 at 12:06
• @ypercube: the second one. Jul 9, 2012 at 12:43
• Added to the card. Thanks very much! Jul 11, 2012 at 1:11
• How does one start to solve this?
– yiyi
Dec 11, 2012 at 8:50
• Using the Risch-Norman algorithm, the antiderivative is not elementary. Maple doesn't find a closed-form antiderivative. Oct 30, 2013 at 14:31

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.

• @PeterTamaroff Thanks. Jul 9, 2012 at 3:33
• Out of curiosity, why the downvote? Jul 9, 2012 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? Jul 9, 2012 at 11:23
• @JoachimSauer That was actually a typo on my part. $108$ was meant to be $118$. Jul 11, 2012 at 3:07
• @حكيمالفيلسوفالضائع It's only a conjecture?
– MT_
May 2, 2014 at 21:36

There's also

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

• The 118 is a bit too obvious, assuming it's not just coincidence that 118/2 = 59 Jul 9, 2012 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, 2012 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.... Jan 12, 2021 at 4:43

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...

• Waiting on your edit as promised.... :-) Jun 13, 2013 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. Jun 13, 2013 at 5:24
• i.imgur.com/8PfsQ.png May 12, 2014 at 0:09
• Almost 8 years later. You still haven't posted the solution. Please... Oct 10, 2021 at 2:15
• Still waiting for the solution! :D Jan 22, 2023 at 2:01