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

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.

## 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. – Surya Prakash · 1 year, 3 months ago

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– Ishan Singh · 1 year, 3 months ago

How will you evaluate the resulting summation?Log in to reply

– Aditya Kumar · 1 year, 3 months ago

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

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– Aditya Kumar · 1 year, 3 months ago

I'll not give up on that problem. I have to solve it first.Log in to reply

– Ishan Singh · 1 year, 3 months ago

Integrate the generating function of harmonic number.Log in to reply

– Surya Prakash · 1 year, 3 months ago

I made it on my own and i dont know whether it is already there or notLog in to reply

– Aditya Kumar · 1 year, 3 months ago

I was asking about the expansion.Log in to reply

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

\[\displaystyle \sum_{r,s,t\geq 0} \frac{H_r(-1)^r}{(r+1)(t+1)(r+s+t+3)^2}\] – Aman Rajput · 1 year, 3 months ago

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@Pi Han Goh add it to you set – Aditya Kumar · 1 year, 3 months ago

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Added! Liked + Reshared! – Pi Han Goh · 1 year, 3 months ago

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– Aditya Kumar · 1 year, 3 months ago

Thanks. Do try it.Log in to reply

Hint :Convert into derivative of Beta function. – Ishan Singh · 1 year, 3 months agoLog in to reply

Which is imaginary, but the closed form ain't contain any imaginary term. – Harsh Shrivastava · 1 year, 3 months ago

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this – Ishan Singh · 1 year, 3 months ago

No. Take limit. For exampleLog in to reply

– Surya Prakash · 1 year, 3 months ago

can u clearly explain how u applied the limit there?Log in to reply

– Ishan Singh · 1 year, 3 months ago

That will take \(2 -3\) pages. I may post it when I'm free.Log in to reply

– Ishan Singh · 1 year, 3 months ago

You have to do something with the \(\ln^2 (1+x)\) term and afterwards it gets converted into the limit. Another method is to use generating function of Harmonic numbers.Log in to reply

You are referring this \(\sum_{n=1}^{\infty} H_{n} x^{n} = \dfrac{-ln(1-x)}{1-x}\) no? – Harsh Shrivastava · 1 year, 3 months ago

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– Ishan Singh · 1 year, 3 months ago

Yes. I'm referring to that generating function.Log in to reply

– Harsh Shrivastava · 1 year, 3 months ago

Kk I'll give another shot at Feynman's way.Log in to reply

@Mark Hennings could you help us out? – Aditya Kumar · 1 year, 3 months ago

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I got it. Give me one hour of time. – Surya Prakash · 1 year, 3 months ago

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