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Wolfram says it is approximately $-0.65111415588...$ but I am hoping for an exact solution. Here is my attempt:

$$\int\limits_0^{\ln{2}}\ln\left(e^{x}+1\right)\ln\left(e^{x}-1\right)dx$$ $$\int\limits_0^{\ln{2}}\ln\left(u\right)\ln\left(e^{x}-1\right)\frac{du}{e^x}$$ $$\int\limits_2^3 e^{-x}\ln\left(u\right)\ln\left(u-2\right)du$$ $$\int\limits_2^3 e^{-\ln\left(u-1\right)}\ln\left(u\right)\ln\left(u-2\right)du$$ $$\int\limits_2^3 e^{-\ln\left(u-1\right)}\ln\left(u\right)\ln\left(u-2\right)du$$ $$\int\limits_2^3 \frac{\ln\left(u\right)\ln\left(u-2\right)}{u-1}du$$

This is where I got stuck. I am thinking a series expansion might help but I am not sure.

Dylan Levine
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  • I wrote "exact solution" instead of "closed form solution" because I do not expect it to have a closed form. I was hoping there might be a representation using special functions. – Dylan Levine Jan 24 '24 at 14:10
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    I have no reason to think it has an exact solution. I just came up with a problem as a challenge for myself and couldn't solve it, so I posted it here. – Dylan Levine Jan 24 '24 at 14:13
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    Integrals of this form can essentially always be given in terms of polylogarithm functions. The theory of multiple zeta values trivialises such logarithmic integrals. – KStarGamer Jan 24 '24 at 14:46

1 Answers1

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Substitute $t=e^x-1$ $$\int\limits_0^{\ln{2}}\ln\left(e^{x}+1\right)\ln\left(e^{x}-1\right)dx=\int_0^1 \frac{\ln(t+2)\ln t}{t+1}\ dt= -\frac{13}{24}\zeta(3)$$ where the last step utilizes the result.

Quanto
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