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I want a closed form for the following integral $$ \int_0^1 \frac1x\;\ln\left(\frac {\ln\left(\frac{1-x}{2}\right)}{\ln\left(\frac{x+1}{2}\right)}\right)\, \mathrm{d}x $$

An integration by parts attempt on my end led to the following integral $$ \int_0^1 \frac1x \;\ln\left(\frac{\ln\left(\frac{1-x}{2}\right)}{\ln\left(\frac{x+1}{2}\right)}\right) \, \mathrm{d}x = \int_0^1 \ln(x)\left(\frac{1}{(1-x)\ln\left(\frac{1-x}{2}\right)} + \frac{1}{(1+x)\ln\left(\frac{1+x}{2}\right)}\right) \, \mathrm{d}x $$

I have seen variations of these integrals, and usually they are solved by Feynman's trick. I think the first one admits to a nice result, but I am not sure on how to handle the second one. If anyone has any ideas please let me know, there might be a substitution I am not seeing?

Apologies for the small integral sign for such a big integrand, I am unsure on how to make them bigger.

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Jessie Christian
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    Unsure if this will help, but if you make the substitution: $$ t =\frac{1 + x}{2}$$ Then the Integral becomes: $$ 2\int_{\frac{1}{2}}^1 \frac{1}{2t - 1}\ln\left( \frac{\ln\left(1 - t\right)}{\ln\left(t\right)}\right):dt $$ – David Galea Apr 16 '24 at 03:00
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    What is the source of the integral? – Martin.s Apr 17 '24 at 02:31
  • Hello, so I was gonna post more of my work as I went on, but this problem originated from me looking deeper into the following integral. https://math.stackexchange.com/questions/915175/prove-int-0-pi-2-ln-leftx2-ln-cos-x2-right-dx-pi-ln-ln2

    If you look at the second answer, we reasoned that the integrand is the real part of some other complex logarithm, and then used a power series argument for his solution. This integral comes about if you try to look for the Imaginary part of that complex logarithm, and then integrate over that function.

     – Jessie Christian Apr 17 '24 at 04:34
  • The imaginary part is given here through some of my work, but if you integrate over that function, you get an identical answer to the given integral here, up to a factor of -1/2 anyways. I’ll add more as I discover more, but was able to at least show that much. – Jessie Christian Apr 17 '24 at 04:41
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    any further progress? – Martin.s May 21 '24 at 04:31
  • @Martin.s Hello, the most I was able to arrive at was an integral representation of part of the solution. I will try to post what I have found in the coming weeks, though I will need to consult my notes. – Jessie Christian Jun 11 '24 at 01:11

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