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How we can show the following definite integral on $[0,1]$?

$\begin{eqnarray}‎ ‎ \int _0^1\left(\ln \frac{x}{1-x}\right)^kdx=(2^k-2)\pi ^k|B_k|, ‎\end{eqnarray}‎$

where $B_k$ are the $k$-th Bernoulli numbers, $k=1,2,\ldots$, respectively.

Note that the Bernoulli numbers $B_k$'s are a sequence of signed rational numbers. For every odd $k$ other than $1$, $B_k=0$. For every even $k$ other than $0$, $B_k$ is negative if $k$ is divisible by $4$ and positive otherwise. The first few Bernoulli numbers for $k=0,1,2,3,4,5,6,7,8,9,10$ are

$B_k=1,-\frac{1}{2}, \frac{1}{6}, 0,-\frac{1}{30},0,\frac{1}{42},0,-\frac{1}{30},0,\frac{5}{66}$.

M. Raha
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1 Answers1

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Performing the change of integration variable $\frac{x}{{1 - x}} = e^{t}$ from $x$ to $t$ turns your integral into $$ \int_{-\infty}^{ + \infty } {\frac{{t^k e^t }}{{(e^t + 1)^2 }}dt} . $$ If $k$ is positive and even, the integrand is even. Hence, by http://dlmf.nist.gov/25.5.E4 and http://dlmf.nist.gov/25.6.E2, we find $$ \int_{-\infty}^{ + \infty } {\frac{{t^k e^t }}{{(e^t + 1)^2 }}dt} = 2\int_{0}^{ + \infty } {\frac{{t^k e^t }}{{(e^t + 1)^2 }}dt} = 2(1 - 2^{1 - k} )k!\zeta (k) = (2^k-2)\pi^k |B_k|. $$ For positive odd $k$, the integrand is an odd function, whence the integral is $0$. Thus your claim is indeed true for all positive integer $k$.

Gary
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