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Prove that $$\int_0^1 \psi{(x) \sin(2 n \pi x)} \space\mathrm{dx}=-\frac{\pi}{2}, \space n\ge1$$ where $\psi(x)$ - digamma function

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What is $\psi(x)$? – Qiaochu Yuan Dec 28 '12 at 9:44
The claim is clearly wrong for $n=0$. – Hagen von Eitzen Dec 28 '12 at 9:53
An interesting commentary for the Riemann-Lebesgue lemma! – GEdgar Dec 28 '12 at 14:13
up vote 10 down vote accepted

By the log-differentiation of the Euler's reflection formula, we have

$$ \psi_0(x) - \psi_0(1-x) = -\pi \cot (\pi x). $$

Thus we have

\begin{align*} \int_{0}^{1}\psi_0(x) \sin (2\pi n x) \, dx &= \frac{1}{2}\int_{0}^{1}\psi_0(x) \sin (2\pi n x) \, dx - \frac{1}{2}\int_{0}^{1}\psi_0(1-x) \sin (2\pi n x) \, dx \\ &= -\frac{\pi}{2} \int_{0}^{1} \frac{\sin (2\pi n x)}{\sin (\pi x)} \, \cos (\pi x) \, dx \\ &= -\frac{1}{2} \int_{0}^{\pi} \frac{\sin (2 n \theta)}{\sin \theta} \, \cos \theta \, d\theta \\ &= - \int_{0}^{\frac{\pi}{2}} \frac{\sin (2 n \theta)}{\sin \theta} \, \cos \theta \, d\theta. \end{align*}

Now the rest follows by my blog posting.

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@Chris'ssister, thank you. :) – Sangchul Lee Dec 28 '12 at 10:04
@sos440 By the way, cool blog you have out there. – Sasha Dec 28 '12 at 14:36
@sos440: I like your use of the reflection formula. (+1) – user26872 Dec 29 '12 at 20:08

Here's another approach using the integral representation for $\psi$. We assume $n$ is an integer greater than or equal to one. Then $$\begin{eqnarray*} \int_0^1 dx\, \sin(2n\pi x) \psi(x) &=& \int_0^1 dx\, \sin(2n\pi x) \int_0^\infty dt\, \left( \frac{e^{-t}}{t} - \frac{e^{-x t}}{1-e^{-t}} \right) \\ &=& \int_0^\infty dt\, \left( \frac{e^{-t}}{t} \int_0^1 dx\, \sin(2n\pi x) - \frac{1}{1-e^{-t}} \int_0^1 dx\, \sin(2n\pi x)e^{-x t} \right). \end{eqnarray*}$$ But $\int_0^1 dx\, \sin(2n\pi x) = 0$ and $$\int_0^1 dx\, \sin(2n\pi x)e^{-x t} = \frac{2n\pi}{t^2+4n^2\pi^2}(1-e^{-t}).$$ (Details for the second integral can be given if necessary.) Therefore $$\begin{eqnarray*} \int_0^1 dx\, \sin(2n\pi x) \psi(x) &=& -\int_0^\infty dt\, \frac{2n\pi}{t^2+4n^2\pi^2} \\ &=& -\frac{\pi}{2}. \end{eqnarray*}$$

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${(\text{precious})}^{\text{precious}}$ (+1) :-) Your way is very short and easy. Thanks! – user 1618033 Dec 29 '12 at 20:11
@Chris'ssister: Glad to help. I had not seen this interesting integral before. (+1) – user26872 Dec 29 '12 at 21:08

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