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How to compute the integral $$I(y):=\int_0^\infty \frac{\ln (1+x^2y^2)}{1+x^2} dx$$

$$ \frac{d I}{dy}= \int_0^\infty \frac{2y x^2}{(1+x^2)(1+x^2y^2)} dx =\frac{\pi}{1+y} $$ Actually, I kind of being stuck at the final step in the beginning. Then, I find that I can use formula for the following indefinite integral $$ \int \frac{1}{a x^2 +b x+c} dx $$ or the Residue Integration Method.

Harry Peter
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Yuhang
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1 Answers1

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Hint would be assume y as variable here and use leibnitz rule to calculate $\frac {dI}{dy} $ this converts integral into $\int _0 ^{\infty} \frac {2yx^2}{(1+x^2y^2)(1+x^2)} $ this can be handled using partial fractions This is an easily integrable form whose answer is $\frac {(\pi)(y-1)}{y^2-1}=\frac {\pi}{y+1} $ . Integrate this wrt to $y $ that would be the final answer.

Archis Welankar
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  • Could you explain how you get the result $\frac{\pi}{y+1}$. I get it from the formula $\int_0^\infty \frac{1}{a^2+y^2} dy=\frac{\pi}{a 2}$ – Yuhang Dec 15 '17 at 13:58
  • antiderivative is $\displaystyle \frac{2(\arctan(xy)-y\arctan x)}{1-y^2}$ and $I(0)=0$. – FDP Dec 15 '17 at 14:11