# Tricky Integral

Can one show that

$$\int_{0}^{\infty} \! \frac {1}{1+x^{2}} \frac {x^{a}-x^{b}}{(1+x^{a})(1+x^{b})}~\mathrm{d}x=0 ~~~~~~~~ \forall ~a,b~\in \mathbb{R}.$$

Any hints?

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is it $x^{x^{b}}$ –  user9413 Jun 5 '11 at 11:06
@Chandru: No, it is just $(1+x^b)$. –  night owl Jun 5 '11 at 11:27
@Rasholnikov: Right, but I think it wants it for any general $(a,b) \in \mathbb{R}$. –  night owl Jun 5 '11 at 11:29

## 1 Answer

Yes, one can. Here are some hints, which should be expanded before being called a proof.

Writing $x^a-x^b$ as $(x^a+1)-(x^b+1)$ and simplifying the fraction, one sees that it is enough to show that $I(a)$ does not depend on $a$, with $$I(a)=\int_0^{+\infty}\frac{\mathrm{d}x}{(1+x^2)(1+x^a)}$$ To prove this, one could decompose $I(a)$ as the sum of an integral from $0$ to $1$ and an integral from $1$ to $+\infty$ and use the change of variable $y\leftarrow1/x$ in the latter. One would be left with $$I(a)=\int_0^{1}\frac{\mathrm{d}x}{(1+x^2)(1+x^a)}+\int_0^{1}\frac{y^a\mathrm{d}y}{(1+y^2)(1+y^a)}=\int_0^{1}\frac{\mathrm{d}x}{1+x^2},$$ which is independent on $a$, and this would yield the result.

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+1$\ldots~$marvelous. –  night owl Jun 5 '11 at 11:34
@night owl: thanks for this excessive but nice comment. –  Did Aug 16 '11 at 21:23