Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can one show that, for every $a$ and $b$ in $\mathbb R$,

$$ \int_{0}^{\infty} \! \frac {1}{1+x^{2}} \frac {x^{a}-x^{b}}{(1+x^{a})(1+x^{b})}~\mathrm{d}x=0\ ? $$

Any hints?

share|cite|improve this question
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
up vote 23 down vote accepted

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.

share|cite|improve this answer
+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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.