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My claim is that: $$\int_0^\infty\frac{1}{(1+x^a)(1+x)^2}dx=\frac12$$ regardless of the value of $a$. A couple examples are here and here.

I came across this result while searching cool integrals to evaluate, but I could not tame this one. I know a similar result, that you can find in this answer. It states that: $$\int_0^\infty\frac{1}{(1+x^a)(1+x^2)}dx=\frac{\pi}{4}$$ and it is proved using the substitution $x=\tan u$, however using the same substitution in my integral seems to be pointless. How should one proceed?

Zima
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1 Answers1

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\begin{align} &\int_0^\infty\frac{1}{(1+x^a)(1+x)^2}\overset{x\to\frac1x}{dx}\\ =& \int_0^\infty\frac{1}{(1+x^{-a})(1+x)^2}dx\\ =& \ \frac12 \int_0^\infty\frac{1}{(1+x)^2} \left( \frac{1}{1+x^{a}}+\frac{1}{1+x^{-a}} \right)dx\\ =&\ \frac12 \int_0^\infty\frac{1}{(1+x)^2}dx=\frac12 \end{align}

Quanto
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  • I tried differentiating the integral w.r.t. $a$, and since the integral is just a constant, is should give $0$. But it does not. Why is this the case? – Zima May 15 '23 at 21:03
  • @Zima - It is zero, as shown below. $$I’(a)=-\int_0^\infty \frac{\ln x}{(x^{a/2}+x^{-a/2})^2(1+x)^2}dx=0$$ – Quanto May 15 '23 at 21:10
  • but derivative of $\frac{1}{1+x^c}$ w.r.t. c is $-\frac{x^c\log x}{(1+x^c)^2}$, and when you integrate together with $\frac{1}{(1+x)^2}$ is not $0$ – Zima May 15 '23 at 21:18
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    @Zima - Note that $\frac{x^a}{(1+x^a)^2}= \frac1{(x^{a/2}+ x^{-a/2})^2}$, which leads to $I’(a)=0$ as stated in above comment. – Quanto May 15 '23 at 21:33