Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|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
add comment

1 Answer

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|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
add comment

Your Answer

 
discard

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.