Prove this closed form of 01(ln(1+x))2ln(x)ln(1x)1xdx\displaystyle \int_0^1\frac{(\ln(1+x))^2\ln(x)\ln(1-x)}{1-x} \, dx

01(ln(1+x))2ln(x)ln(1x)1xdx=72ζ(3)(ln2)2π26(ln2)3π22ζ(3)+6ζ(5)π448ln2\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: ζ()\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.

Note by Aditya Kumar
3 years, 7 months ago

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Use this expansion,

ln2(1+x)=r=0Hr(1)rxr+1r+1\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

01xr+1ln(x)ln(1x)1xdx\large \int_{0}^{1} x^{r+1} \dfrac{\ln(x) \ln(1-x)}{1-x} dx

This is derivative of beta function.

Surya Prakash - 3 years, 7 months ago

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How will you evaluate the resulting summation?

Ishan Singh - 3 years, 7 months ago

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

Aditya Kumar - 3 years, 7 months ago

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@Pi Han Goh add it to you set

Aditya Kumar - 3 years, 7 months ago

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Added! Liked + Reshared!

Pi Han Goh - 3 years, 7 months ago

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Thanks. Do try it.

Aditya Kumar - 3 years, 7 months ago

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

r,s,t0Hr(1)r(r+1)(t+1)(r+s+t+3)2\displaystyle \sum_{r,s,t\geq 0} \frac{H_r(-1)^r}{(r+1)(t+1)(r+s+t+3)^2}

Aman Rajput - 3 years, 7 months ago

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@Mark Hennings could you help us out?

Aditya Kumar - 3 years, 7 months ago

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Hint : Convert into derivative of Beta function.

Ishan Singh - 3 years, 7 months ago

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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.

Harsh Shrivastava - 3 years, 7 months ago

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No. Take limit. For example this

Ishan Singh - 3 years, 7 months ago

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@Ishan Singh You have to do something with the ln2(1+x)\ln^2 (1+x) term and afterwards it gets converted into the limit. Another method is to use generating function of Harmonic numbers.

Ishan Singh - 3 years, 7 months ago

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@Ishan Singh Kk I'll give another shot at Feynman's way.

Harsh Shrivastava - 3 years, 7 months ago

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@Ishan Singh I got imaginary term while using that Harmonic relation.

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

Harsh Shrivastava - 3 years, 7 months ago

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@Harsh Shrivastava Yes. I'm referring to that generating function.

Ishan Singh - 3 years, 7 months ago

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@Ishan Singh can u clearly explain how u applied the limit there?

Surya Prakash - 3 years, 7 months ago

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@Surya Prakash That will take 232 -3 pages. I may post it when I'm free.

Ishan Singh - 3 years, 7 months ago

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