How do I Calculate, if possible, in terms of well-known constants the integral :
$\int_{0}^{1}x^{k}\psi(x)dx$ , where $k\geq 3$ is an integer ?
note: $\psi(x)$ is digamma function.
Any help would be greatly appreciated.
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$\begingroup$ Integration by parts works here... $\endgroup$– Feldmann DenisCommented May 30, 2015 at 17:39
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$\begingroup$ i think recursive formula $\endgroup$– salimmath15Commented May 30, 2015 at 17:43
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2 Answers
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This integral has been observed by Donal F. Connor in 2010 (you can find the link here, pg. 94). As far as I know, he found a closed form for the odd case, but I believe the even case is somewhere in that document (don't quote me on that yet). To solve it, as Feldmann Denis notes, use integration by parts.
According to Dr. Connor, we have $$ \int_0^1 x^{2n+1}\psi{(x)} \;dx = \sum_{k=0}^{2n} \left( \begin{array}{c} 2n+1 \\ k \end{array} \right) \big(H_k\zeta{(-k)} + (-1)^{k+1}\zeta'{(-k)}\big)$$
where $H_k$ is the harmonic series, and $\zeta$ is the Riemann zeta function.
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1$\begingroup$ j.Diaz , thank you for your answer but you take only k as odd integer, what about k is even ? $\endgroup$ Commented May 30, 2015 at 18:14
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Hint :
I think this is a closed form of :$\int_{0}^{1}x^{k}\psi(x)dx$ for $ k>2$:
for $ k>2 $:
$\int_{0}^{1}x^{k}\psi(x)dx$=$-log\sqrt{2\pi} +\sum_{j=1}^{k-1}(-1)^{j+1}C_{k}^{j}logA_{j}$
where :$A_{j},j \geq 1$ is the generalized Glaisher-Kinkelin constant.