6

Question:

Prove that $$\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx=\frac{-π^2}{16(2+\sqrt{2})}$$

Using partial fraction,we get

\begin{align} &\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx\\ = &\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1-x}dx+\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1+x}dx+\frac{1}{2}\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx \end{align}

I got first integral as $\frac{-\pi^2}{6}$ and second integral as $\frac{-\pi^2}{12}$

$$\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx=\frac{-\pi^2}{16}+\frac{1}{2}\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx$$

I got stuck with third integral, which seems difficult to evaluate. ${}{}{}$ A note is also written saying that:-
The reader should evaluate the family of integrals ${I_{n}=\int_{0}^{1}\frac {x^{2n}\ln x}{{(1-x^2)}{(1+x^4)^n}}dx{,n} \in N}$ The computation of the first few special values indicates an interesting arithmetic structure of the answer.

How to evaluate integral for $n$?

Quanto
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Paras
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1 Answers1

3

Rewrite the third integral as

$$I=\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx = -\int_{0}^{\infty}\frac{\ln x}{1+x^4}dx $$

Use the general result

$$J(a)= \int_{0}^{\infty}\frac{x^{a-1}}{1+x^m}dx=\frac{\pi}m\csc\frac{\pi a}m $$

and let $m=4$ to obtain $$I= - J’(1)= \frac{\pi^2}{8\sqrt2} $$

Quanto
  • 120,125