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1. APOLOGY

I had previously asked a similar question, but that question was closed due to negligence on my part (too little information, lack of respect for the respondent, etc.). I am very sorry about this. I felt that in the future I must take care to ensure that this does not happen to me when I ask questions.

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2. BACKGROUND

A few months ago, I posted a question about the following integral

$$ \int_0^1 \! \frac{x}{\operatorname{artanh}x} \, dx = \int_0^1 \! \frac{\sinh x}{x\cosh^3 x} \, dx = \frac{7\,\zeta(3)}{\pi^2} $$

where $\operatorname{artanh}x$ is the inverse hyperbolic tangent function. Furthermore, I computed the following integrals

$$ \int_0^1 \!\!\! \int_0^1 \! \frac{\ln\frac1x - \ln\frac1y}{\ln\ln\frac1x - \ln\ln\frac1y} \, dx \, dy = \frac{7\,\zeta(3)}{\pi^2} $$

and obtained exactly the same results as the integrals above. Since the values of both integrals are equal, we get the conclusion that

$$ \int_0^1 \! \frac{x}{\operatorname{artanh} x} \, dx = \int_0^1 \!\!\! \int_0^1 \! \frac{\ln\frac1x - \ln\frac1y}{\ln\ln\frac1x - \ln\ln\frac1y} \, dx \, dy, $$

but according to my friend,

$$ \int_0^1 \!\!\! \int_0^1 \Biggl( \frac{\ln\frac1x - \ln\frac1y}{\ln\ln\frac1x - \ln\ln\frac1y} \Biggr)^{\!n} \, dx \, dy = \frac{\varGamma(n+2)}{2^n} \int_0^1 \biggl( \frac{x}{\operatorname{artanh} x} \biggr)^{\!n} \, dx $$

holds more generally for all positive integers $n$ where $\varGamma$ is the gamma function.

My friend and I have not yet obtained the closed-form of the above integral. Therefore, I believe he has shown this only by deformation of the integral. So, I too would like to show this result only by the transformation of the integrals, without comparing the values of the integrals, but I cannot show it even for a case $n=1$.

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3. MAIN QUESTION

Can you show that

$$ \int_0^1 \!\!\! \int_0^1 \Biggl( \frac{\ln\frac1x - \ln\frac1y}{\ln\ln\frac1x - \ln\ln\frac1y} \Biggr)^{\!n} \, dx \, dy = \frac{\varGamma(n+2)}{2^n} \int_0^1 \biggl( \frac{x}{\operatorname{artanh} x} \biggr)^{\!n} \, dx $$

only with transformation of the integral, without comparing by the value of the integral?

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4. FINALLY

I have been very disrespectful to you all respondents. I am very sorry.

Kei Tojo
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1 Answers1

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Starting from the double integral \begin{equation} I_2=\int_0^1 \int_0^1 \left( \frac{\ln\frac1x - \ln\frac1y}{\ln\ln\frac1x - \ln\ln\frac1y} \right)^{n} \, dx \, dy %= \frac{\Gamma(n+2)}{2^n} \int_0^1 \left( \frac{x}{\operatorname{artanh} x} \right)^{\!n} \, dx \end{equation} and changing $x=e^{-u},y=e^{-v}$ one obtains \begin{equation} I_2=\int_0^\infty \int_0^\infty \left( \frac{u-v}{\ln u - \ln v} \right)^{n}e^{-u-v} \, du \, dv \end{equation} Then, with $u=sv$, \begin{equation} I_2=\int_0^\infty \int_0^\infty v^{n+1}\left( \frac{s-1}{\ln s} \right)^{n}e^{-v(s+1)} \, ds \, dv \end{equation} Performing first the integration over $v$ (assuming Fubini conditions are fulfilled) \begin{align} I2&=\int_0^\infty \left( \frac{s-1}{\ln s} \right)^{n} \, ds\int_0^\infty v^{n+1}e^{-v(s+1)} \, dv\\ &=\Gamma(n+2)\int_0^\infty \left( \frac{s-1}{\ln s} \right)^{n} \frac{ ds}{(s+1)^{n+2}} \end{align} The latter integral can be split into \begin{equation} \int_0^1 \left( \frac{s-1}{\ln s} \right)^{n} \frac{ ds}{(s+1)^{n+2}}+\int_1^\infty \left( \frac{s-1}{\ln s} \right)^{n} \frac{ ds}{(s+1)^{n+2}} \end{equation} Changing $s\rightarrow 1/s$ shows that both intgrals of this sum are equal, thus \begin{equation} I_2=2\Gamma(n+2)\int_0^1 \left( \frac{s-1}{\ln s} \right)^{n} \frac{ ds}{(s+1)^{n+2}} \end{equation} Now, changing $s=(1-x)/(1+x)$ gives \begin{align} I_2&=\Gamma(n+2)\int_0^1\left(\frac x{ \ln\frac {1+x}{1-x}} \right)^n\,dx\\ &=\frac{\Gamma(n+2)}{2^n} \int_0^1 \left( \frac{x}{\operatorname{artanh} x} \right)^{\!n} \, dx \end{align} as expected.

Paul Enta
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