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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.

Note by Aditya Kumar
7 months, 1 week 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. Surya Prakash · 7 months, 1 week ago

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@Surya Prakash How will you evaluate the resulting summation? Ishan Singh · 7 months, 1 week ago

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@Surya Prakash wow nice one! Do you know how to prove it? Aditya Kumar · 7 months, 1 week ago

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Comment deleted 7 months ago

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@Ishan Singh I'll not give up on that problem. I have to solve it first. Aditya Kumar · 7 months, 1 week ago

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@Aditya Kumar Integrate the generating function of harmonic number. Ishan Singh · 7 months, 1 week ago

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@Aditya Kumar I made it on my own and i dont know whether it is already there or not Surya Prakash · 7 months, 1 week ago

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@Surya Prakash I was asking about the expansion. Aditya Kumar · 7 months, 1 week ago

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

\[\displaystyle \sum_{r,s,t\geq 0} \frac{H_r(-1)^r}{(r+1)(t+1)(r+s+t+3)^2}\] Aman Rajput · 7 months, 1 week ago

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@Pi Han Goh add it to you set Aditya Kumar · 7 months, 1 week ago

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@Aditya Kumar Added! Liked + Reshared! Pi Han Goh · 7 months, 1 week ago

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@Pi Han Goh Thanks. Do try it. Aditya Kumar · 7 months, 1 week ago

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Hint : Convert into derivative of Beta function. Ishan Singh · 7 months, 1 week ago

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@Ishan Singh 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 · 7 months, 1 week ago

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@Harsh Shrivastava No. Take limit. For example this Ishan Singh · 7 months, 1 week ago

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@Ishan Singh can u clearly explain how u applied the limit there? Surya Prakash · 7 months, 1 week ago

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@Surya Prakash That will take \(2 -3\) pages. I may post it when I'm free. Ishan Singh · 7 months, 1 week ago

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@Ishan Singh 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. Ishan Singh · 7 months, 1 week ago

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@Ishan Singh 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? Harsh Shrivastava · 7 months, 1 week ago

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@Harsh Shrivastava Yes. I'm referring to that generating function. Ishan Singh · 7 months, 1 week ago

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@Ishan Singh Kk I'll give another shot at Feynman's way. Harsh Shrivastava · 7 months, 1 week ago

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@Mark Hennings could you help us out? Aditya Kumar · 7 months, 1 week ago

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I got it. Give me one hour of time. Surya Prakash · 7 months, 1 week ago

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