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

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

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.

1 year, 9 months ago

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Use 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.

- 1 year, 9 months ago

How will you evaluate the resulting summation?

- 1 year, 9 months ago

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

- 1 year, 9 months ago

Comment deleted Feb 29, 2016

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

- 1 year, 9 months ago

Integrate the generating function of harmonic number.

- 1 year, 9 months ago

I made it on my own and i dont know whether it is already there or not

- 1 year, 9 months ago

- 1 year, 9 months ago

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

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

- 1 year, 8 months ago

@Pi Han Goh add it to you set

- 1 year, 9 months ago

- 1 year, 9 months ago

Thanks. Do try it.

- 1 year, 9 months ago

Hint : Convert into derivative of Beta function.

- 1 year, 9 months ago

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.

- 1 year, 9 months ago

No. Take limit. For example this

- 1 year, 9 months ago

can u clearly explain how u applied the limit there?

- 1 year, 9 months ago

That will take $$2 -3$$ pages. I may post it when I'm free.

- 1 year, 8 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.

- 1 year, 9 months ago

I got imaginary term while using that Harmonic relation.

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

- 1 year, 9 months ago

Yes. I'm referring to that generating function.

- 1 year, 9 months ago

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

- 1 year, 9 months ago

@Mark Hennings could you help us out?

- 1 year, 9 months ago

I got it. Give me one hour of time.

- 1 year, 9 months ago