\[\int_0^1\frac{(\ln(1+x))^2\ln(x)\ln(1-x)}{1-x} \, dx=\dfrac{7}{2}\zeta(3){(\ln 2)^2}-\dfrac{\pi^2}{6}{(\ln 2)^3}-\dfrac{\pi^2}{2}\zeta(3)+{6}\zeta(5)-\dfrac{\pi^4}{48}\ln2\]

Prove that the equation above is true.

**Notation**: \(\zeta(\cdot) \) denotes the Riemann Zeta function.

This was found in another mathematics form and it was unanswered there.

This is a part of the set Formidable Series and Integrals.

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## Comments

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TopNewestUse this expansion,

\[\large \ln ^2 (1+x) = \sum_{r=0}^{\infty} \dfrac{H_{r} (-1)^r x^{r+1}}{r+1}\]

Then it will be left to evaluate the integral

\[\large \int_{0}^{1} x^{r+1} \dfrac{\ln(x) \ln(1-x)}{1-x} dx\]

This is derivative of beta function.

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How will you evaluate the resulting summation?

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wow nice one! Do you know how to prove it?

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@Pi Han Goh add it to you set

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Added! Liked + Reshared!

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Thanks. Do try it.

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Now @Ishan Singh can solve this.. Integral can be written as:

\[\displaystyle \sum_{r,s,t\geq 0} \frac{H_r(-1)^r}{(r+1)(t+1)(r+s+t+3)^2}\]

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@Mark Hennings could you help us out?

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Hint :Convert into derivative of Beta function.Log in to reply

Yup I did that but at some point, I have to take natural logarithm of (-1).

Which is imaginary, but the closed form ain't contain any imaginary term.

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No. Take limit. For example this

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You are referring this \(\sum_{n=1}^{\infty} H_{n} x^{n} = \dfrac{-ln(1-x)}{1-x}\) no?

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