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My friend gave me the following integral

$$ \int_0^1 \frac{\ln(1-x)}{1+s^2 x^2} \, dx = \frac{\arctan s \ln(1+s^2)}{2s} - \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \, s^{2n} $$

which is valid for $s\in(-1,1)$. I have experimented numerically and this seems to hold true.

I thought that I could prove the identity easily by using the Taylor series expansion

$$ \ln(1-x) = - \sum_{n=1}^\infty \frac{x^n}{n} ,\quad \frac1{1+s^2x^2} = \sum_{n=0}^\infty \, (-1)^n \, (sx)^{2n} $$

or using the formula

$$ \frac{d}{dx} \, \arctan sx = \frac{s}{1+s^2 x^2} $$

but using either method, I was unable to prove. In other words, one would think that this integral could be easily solved by using basic techniques such as Taylor series expansion, but in fact none of these techniques could be successfully applied. This integral is likely to be more difficult than it appears.

Can you show that

$$ \int_0^1 \frac{\ln(1-x)}{1+s^2 x^2} \, dx = \frac{\arctan s \ln(1+s^2)}{2s} - \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \, s^{2n}? $$

Zacky
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Kei Tojo
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  • Your integral is essentially calculated in this question (look at the Update + the comments). You just need to deal with $\int_0^1 \frac{\ln x}{1+s^2x^2}dx$ separately. – Zacky Jan 26 '24 at 15:33

1 Answers1

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Split the integral as follows $$ \int_0^1 \frac{\ln(1-x)}{1+s^2 x^2} \, dx = \int_0^1 \frac{\ln\frac{1-x}x}{1+s^2 x^2} \, dx + \int_0^1 \frac{\ln x}{1+s^2 x^2} \, dx $$ where

\begin{align} &\int_0^1 \frac{\ln\frac{1-x}x}{1+s^2 x^2} \, \overset{y=\frac{1-x}x }{dx }\\ =&\int_0^\infty \frac{\ln y}{(y+1)^2+s^2}\overset{y\to \frac{1+s^2}y}{dy} = \int_0^\infty \frac{\ln(1+s^2)-\ln y}{(y+1)^2+s^2}dy\\ = &\ \frac12 \int_0^\infty \frac{\ln(1+s^2)}{(y+1)^2+s^2}\ dy = \frac1{2s}\ln(1+s^2)\tan^{-1}s \\ \\ &\int_0^1 \frac{\ln x}{1+s^2 x^2} \, dx \\ =& \int_0^1 \ln x \ \sum_{n=0}^\infty(-)^n (s x)^{2n}\ dx\\ = & \sum_{n=0}^\infty(-1)^n s^{2n}\int_0^1 x^{2n}\ln x\ dx=-\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \, s^{2n} \end{align}

Quanto
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