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Double integrals

$$\int_0^1\int_0^1\frac{[-\ln(x)]^s}{1-xy}dxdy=\frac{\zeta(s+2)}{\Gamma(s+2)} \tag1$$

$$\int_0^1\int_0^1\frac{[-\ln(xy)]^s}{1-xy}dxdy=\zeta(s+2)\Gamma(s+2) \tag2$$

Where $\sum_{n=0}^{\infty}\frac{1}{(n+1)^s}=\zeta(s)$, valid for $\Re(s)>1$ and $\Gamma(n+1)=n!$ valid for all non-negative integers and rational arguments.

I came across these two double integrals during the time I was on Wolfram integrator, was trying to search for something. It didn't gave me the closed form, just the numerical values and the rest I had to find the closed form base on these values.

I don't know how to prove these integrals, can somebody show me how to prove it, so I can learn from it, so next time I can independently do it myself.

I hope these closed form are correct.

P.s

Please, try not to miss too many steps. Thank you!

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  • $\begingroup$ you want to prove that $\int_0^1\int_0^1\frac{[-\ln(x)]^s}{1-xy}dxdy=\frac{\zeta(s+2)}{\Gamma(s+2)}$ ? $\endgroup$
    – reuns
    Commented May 8, 2016 at 20:39
  • $\begingroup$ Yes, please, can you help me? $\endgroup$
    – user335850
    Commented May 8, 2016 at 21:15

2 Answers 2

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For the first integral, we can write

$$\begin{align} I(s)&=\int_0^1\int_0^1 \frac{\left(-\log(x)\right)^s}{1-xy}\,dx\,dy\\\\ &=\sum_{n=0}^\infty\left(\int_0^1 y^n \,dy \int_0^1 x^n \log^s(1/x)\,dx \right)\\\\ &=\sum_{n=0}^\infty \frac{1}{n+1}\int_0^1 x^n \log^s(1/x)\,dx\\\\ &=\sum_{n=0}^\infty \frac{1}{n+1} \int_0^\infty t^s\,e^{-(n+1)t}\,dt\\\\ &=\sum_{n=0}^\infty \frac{1}{(n+1)^{s+2}}\int_0^\infty t^s\,e^{-t}\,dt\\\\ &=\zeta(s+2)\Gamma(s+1) \end{align}$$


For the second integral, we can write

$$\begin{align} J(s)&=\int_0^1\int_0^1 \frac{\left(-\log(xy)\right)^s}{1-xy}\,dx\,dy\\\\ &=\sum_{n=1}^\infty \int_0^1 \frac1y \int_0^y t^n\log^s(1/t)\,dt \,dy\\\\ &=\sum_{n=1}^\infty \int_0^1 t^n\log^s(1/t) \int_t^1 \frac1y \,dy \,dt\\\\ &=\sum_{n=1}^\infty \int_0^1 t^n\log^{s+1}(1/t)\,dt\\\\ &=\sum_{n=1}^\infty \int_0^1 e^{-(n+1)u}u^{s+1}\,du\\\\ &=\sum_{n=1}^\infty \frac{1}{(n+1)^{s+2}}\int_0^1 e^{-u}u^{s+1}\,du\\\\ &=\zeta(s+2)\Gamma(s+2) \end{align}$$

as was to be shown!

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  • $\begingroup$ (+1). Is my first stated integral was wrong or there is a type error? $\endgroup$
    – user334593
    Commented May 9, 2016 at 4:12
  • $\begingroup$ The first integral in the OP is incorrect. $\endgroup$
    – Mark Viola
    Commented May 9, 2016 at 4:27
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I can't comment.

See this article:

http://arxiv.org/pdf/math/0506319.pdf

Correlaries 3.1 and 3.2 have your answers

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  • $\begingroup$ Interesting, I only saw one of the answer. Is there a simple way of proving this integral? $\endgroup$
    – user335850
    Commented May 8, 2016 at 21:16
  • $\begingroup$ I edited, pisquare $\endgroup$
    – user338046
    Commented May 8, 2016 at 21:29
  • $\begingroup$ Thank you. I was hoping there is a simple way of integrating these integrals. $\endgroup$
    – user335850
    Commented May 8, 2016 at 21:34
  • $\begingroup$ Yes, see the entire article is using the formulas at the top of the article and then plugging in convinient constants for interesting formulas $\endgroup$
    – user338046
    Commented May 8, 2016 at 21:36

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