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$$\int_0^1 \int_0^1 \frac{1-x}{1-xy}(-\log(xy))^s \, dx \, dy=\Gamma(s+2)\left(\zeta(s+2) -\frac{1}{s+1}\right) \quad \Re(s)>-2$$

How to prove this identity?

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  • $\begingroup$ What is "s" on the LHS? $\endgroup$ – Marra Feb 24 '13 at 13:38
  • $\begingroup$ $s$ is a complex number. $\endgroup$ – Anthony Feb 24 '13 at 13:40
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We will need the result

$$ \int_{0}^{\infty}\frac{t^{s-1}}{e^{t}-1}=\Gamma(s)\zeta(s).$$

We first Write the integral in consideration in the form $$ \int_0^1 \int_0^1 \frac{1-x}{1-xy}(-\ln(xy))^sdx dy = $$

$$ \int_0^1 \int_0^1 \frac{(-\log(xy))^s}{1-xy} dx dy - \int_0^1 \int_0^1 \frac{x(-\log(xy))^s}{1-xy} dx dy. $$

We start with the first integral. Using the change of variables $u=x y$ and then interchanging the order of integration yields

$$ \int_0^1 \int_0^1 \frac{(-\ln(xy))^s}{1-xy} dx dy = \int_0^1 \int_0^y \frac{(-\log(u))^s}{1-u} \frac{du}{y} dy $$

$$= \int_0^1 \frac{(-\ln(u))^s}{1-u} \int_u^1\frac{du}{y} dy du = \int_0^1 \frac{(-\ln(u))^{s+1}}{1-u} du. $$

Using the change of variables $ \ln u = -t $, we get

$$ = \int_0^1 \frac{(-\ln(u))^{s+1}}{1-u} du = - \int_{0}^{\infty}\frac{t^{s+1}}{e^{t}-1}=-\Gamma(s+2)\zeta(s+2). $$

Now, you should be able to finish the problem following the above techniques. I have not worked out the rest of the problem yet.

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I've met a similar exercise, expanding the denominator as a geometric series (which is legal, because $0<x,y<1$ helped me a lot.

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