Prove that
$$\int_0^{\infty} \frac{x^{2n+1}}{e^{\pi x}-1}\mathrm{d}x \in \mathbb{Q}, \forall \,n\in\mathbb{N}$$
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This looks like one of Ramanujan’s integrals! – Haskell Curry Feb 21 '13 at 19:11
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I suggest you put in some serious effort on $n=1,$ then see what happens with induction. May diverge for $n=0,$ a pity. Meanwhile, where did you get this problem? – Will Jagy Feb 21 '13 at 19:14
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Yet another great question of yours. – Julien Feb 21 '13 at 19:20
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It's alright for $n=0,$ the denominator is $\pi x + more$ – Will Jagy Feb 21 '13 at 19:20
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@WillJagy: for $n=0$ Mathematica tells us that the value of the integral is $1/6$. – user 1591719 Feb 21 '13 at 19:24
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@julien: I'm glad you like it. :) – user 1591719 Feb 21 '13 at 19:25
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If you make the change of variables $y=\pi x$ and use the result you get the value of the integral directly. – Mhenni Benghorbal Feb 21 '13 at 19:40
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@MhenniBenghorbal: that's good point! Thanks! – user 1591719 Feb 21 '13 at 19:54
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@Chris'ssisterandpals: You are welcome. – Mhenni Benghorbal Feb 21 '13 at 20:02
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@Will Jagy: the problem is created by Chris. Actually, he experimentally noticed this strange behaviour of the integral and asked me to come up with a solution, but I couldn't do it. – user 1591719 Feb 22 '13 at 09:39
2 Answers
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Rewrite the integral as
$$\begin{align}\int_0^{\infty} dx \: x^{2 n+1} e^{-\pi x} \sum_{k=0}^{\infty} e^{- k \pi x} &= \sum_{k=0}^{\infty} \int_0^{\infty} dx \: x^{2 n+1} e^{-(k+1) \pi x}\\ &= \frac{(2 n+1)!}{\pi^{2 n+2}} \zeta(2 n+2)\end{align}$$
(in the last step, I evaluated the integral and the sum produced a Riemann zeta). We know that, for a zeta of an even integer is
$$\zeta(2 n+2) = (-1)^n \frac{B_{2 n+2} (2 \pi)^{2 n+2}}{2 (2 n+2)!}$$
where $B_k$ is the $k$-th Bernoulli number and is rational. Putting the last two equations together, it is clear that the sought integral is rational for all natural numbers $n$.
Haskell Curry
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Ron Gordon
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There's always another way to solve these problems. This is just the one that came to mind right now. To be honest, I cannot think of another way, but it surely exists. – Ron Gordon Feb 21 '13 at 19:27
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@rlgordonma: I just put a hyphen between ‘$ k $’ and ‘th’ to improve readability. :) – Haskell Curry Feb 21 '13 at 19:44
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@HaskellCurry: not how I normally write things, but OK, thanks for reading that closely. – Ron Gordon Feb 21 '13 at 19:49
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This is an supplement of rlgordonma's answer if one doesn't want to use Bernoulli's numbers.
Start with Euler's infinite product expansion of $\sin(x)$: $$\sin(x) = x \prod_{k=1}^{\infty} \left( 1 - (\frac{x}{k\pi})^2\right)$$ We have for $|x| < \pi$, $$ \log \frac{\sin(x)}{x} = \sum_{k=1}^{\infty} \log( 1 - (\frac{x}{k\pi})^2 ) = - \sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n}(\frac{x}{k\pi})^{2n}\tag{*}$$ Since the double series converges absolutely for $|x| < \pi$, we can switch the order of summation: $$(*) = - \sum_{n=1}^{\infty}\frac{1}{n}\sum_{k=1}^{\infty}(\frac{x}{k\pi})^{2n} = -\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n}(\frac{x}{\pi})^{2n}\tag{**} $$ We can also differentiate the result term by term to get: $$\frac{\cos(x)}{\sin(x)} - \frac{1}{x} = -2 \sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}} x^{2n-1}$$ In the Taylor expansion of $\sin(x)$ and $\cos(x)$, their coefficients are all rational numbers. This mean when we expand L.H.S as a power series in $x$, all its coefficients are rational numbers too. Compare it with those in R.H.S, we get $\frac{\zeta(2n)}{\pi^{2n}}$ are rational numbers for all $n$.
achille hui
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