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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!

  • 64
    Now let's just hope that your father doesn't read math.stackexchange.com...2012-07-09
  • 2
    Wiki-hammered as this is a bit list with a subjective (as hard as possible) answer.2012-07-09
  • 9
    Yeah... or your dad might know right away that the solution evaluates to 59... on his birthday... when he just turned 59.2012-07-10
  • 2
    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 59.2013-12-19

5 Answers 5

21

There's also

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

  • 6
    The 118 is a bit too obvious, assuming it's not just coincidence that 118/2 = 592012-07-09
  • 6
    @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.2012-07-09
120

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

  • 31
    With added benefit of having very few arbitrary-looking constants.2012-07-09
  • 0
    @Alex: should be even better in 2 years! :-)2012-07-09
  • 8
    I like this Easy to read, no crazy exponents yet nontrivial solution.2012-07-09
  • 0
    Wolfram Alpha timed out trying to evaluate it.2013-03-14
  • 1
    @Joe: hint for fast evaluation : use the generating function for [Euler numbers](http://en.wikipedia.org/wiki/Euler_number).2013-03-15
  • 3
    @Raymond 59 = 61 - 2?2013-03-15
  • 0
    @Joe: yes easy isn't it? :-) (and that's why it will be simpler in 2014!)2013-03-15
  • 0
    @RaymondManzoni There should be a lemma I am missing, but why $\int t^{2n}/cosh(t \pi/2) = |E_{2n}|$ ?2013-06-03
  • 0
    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 :2013-06-04
  • 0
    $$I(2n)=(2n)!\;\beta(2n+1)=(-1)^n\;E_{2n}\;\left(\frac {\pi}2\right)^{2n+1}$$ 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).2013-06-04
  • 0
    @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 $\lim_{t\to 0}$.2013-06-04
  • 0
    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 $E_{2n}$.2013-06-04
84

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

  • 6
    Is that $$\ln(x^2)$$ or $$(\ln x)^2$$ ?2012-07-09
  • 4
    @ypercube: the second one.2012-07-09
  • 2
    Added to the card. Thanks very much!2012-07-11
  • 10
    How does one start to solve this?2012-12-11
  • 0
    Does the indefinite integral of the integrand give a closed-form solution?2013-10-30
  • 2
    Using the Risch-Norman algorithm, the antiderivative is not elementary. Maple doesn't find a closed-form antiderivative.2013-10-30
41

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.

  • 0
    @PeterTamaroff Thanks.2012-07-09
  • 6
    Out of curiosity, why the downvote?2012-07-09
  • 8
    @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?2012-07-09
  • 1
    @JoachimSauer That was actually a typo on my part. $108$ was meant to be $118$.2012-07-11
  • 0
    @حكيمالفيلسوفالضائع It's only a conjecture?2014-05-02
  • 0
    @MichaelT After some research I found that it isn't.2014-05-02
23

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...

  • 4
    Waiting on your edit as promised.... `:-)`2013-06-13
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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.2013-06-13
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    http://i.imgur.com/8PfsQ.png2014-05-12