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.

#Calculus

Easy Math Editor

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Comments

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 - 5 years, 3 months ago

How will you evaluate the resulting summation?

Ishan Singh - 5 years, 3 months ago

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

Aditya Kumar - 5 years, 3 months ago

@Pi Han Goh add it to you set

Aditya Kumar - 5 years, 3 months ago

Added! Liked + Reshared!

Pi Han Goh - 5 years, 3 months ago

Thanks. Do try it.

Aditya Kumar - 5 years, 3 months ago

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

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

Aman Rajput - 5 years, 3 months ago

@Mark Hennings could you help us out?

Aditya Kumar - 5 years, 3 months ago

Hint : Convert into derivative of Beta function.

Ishan Singh - 5 years, 3 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.

Harsh Shrivastava - 5 years, 3 months ago

No. Take limit. For example this

Ishan Singh - 5 years, 3 months ago

@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 - 5 years, 3 months ago

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

Harsh Shrivastava - 5 years, 3 months ago

@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 - 5 years, 3 months ago

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

Ishan Singh - 5 years, 3 months ago

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

Surya Prakash - 5 years, 3 months ago

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

Ishan Singh - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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

Comments

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