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.

evaluate $\int_0^\infty \dfrac{dx}{1+x^4}$using $\int_0^\infty \dfrac{u^{p-1}}{1+u} du = \dfrac{\pi}{\sin( \pi p)}$. I am having trouble finding what is $p$. I set $u = x^4$, I figure $du = 4x^3 dx$, I am unsure though how to find $p$ though. Could someone tell me what I am missing? Thanks.

share|improve this question
2  
You may want to write $x = u^{1/4}$. –  sos440 Feb 12 '13 at 7:51
    
Are you sure you have to do it this way? If you can evaluate the second integral by residues, surely you can evaluate the first by residues. –  Potato Feb 12 '13 at 7:51
    
@Potato The point of the problem is to do it using this, for $0 < p < 1$ –  user1535776 Feb 12 '13 at 7:57

1 Answer 1

up vote 5 down vote accepted

If $u=x^4$ and $du=4x^3 dx$ then: $$dx = \frac{1}{4x^3} du = \frac{1}{4{(x^4)}^{3/4}} du = \frac{1}{4{u}^{3/4}} du$$ So your integral is:

$$\int_0^\infty \dfrac{dx}{1+x^4} = \frac{1}{4}\int_0^\infty \frac{u^{-3/4}du}{1+u} = \frac{\pi}{4\sin(\pi/4)}=\frac{\pi }{2 \sqrt{2}}$$

share|improve this answer

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.