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I want to compute $\displaystyle \int^{\infty}_{0}\frac{x}{x^4+1}dx$ using the residue theorem.

The poles in the upper half plane are:

location: $\large e^{\frac{\pi i}4}$, order: 1, residue: $\large\frac{1}{4}e^{\frac{3\pi i}2}$
location: $\large e^{\frac{3\pi i}4}$, order: 1, residue: $\large \frac{1}{4}e^{\frac{\pi i}2}$

The problem is that the integral from $-\infty$ to $\infty$ vanishes for symmetry reasons, so I cannot apply the standard approach of putting the half of a 1-sphere on top of the real axis and letting its radius go to infinity. If x was replaced with $x^2$ for instance, I could just divide the result by two. Is there another way of contour integration to evaluate the upper expression?

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Try the contour $\Gamma_R = \gamma_1+\gamma_2+\gamma_3$ where $\gamma_1(t) = t, 0\leq t\leq R$, $\gamma_2(\theta) = Re^{i\theta}, 0\leq\theta\leq \frac{\pi}3$ and $\gamma_3(t) = (R-t)e^{i\frac{\pi}3}, 0\leq t\leq R$. –  Davide Giraudo Jul 9 '11 at 18:52
    
Can't you substitute $u=x^2$ first? –  Emre Jul 9 '11 at 18:52
1  
I think you mean "improper integral," not "indefinite integral." –  Corey Jul 9 '11 at 19:06
    
@Corey: Yes, of course; My undergraduate courses are not yet in english, so I still have to learn the correct terms.. @girdav: Thanks, I might look into that after I went through math.stackexchange.com/questions/44928/… –  Eli Jul 9 '11 at 19:10
    
See here math.stackexchange.com/questions/60443/… –  Pablo Sep 12 '11 at 21:32

1 Answer 1

up vote 3 down vote accepted

Those integrals were discussed by us in detail already.

To be more explicit you are asking for the special case of Interesting integral formula for $m=2$, $n=4$ and $a=1$. Just directly plugging in those values in the proof you get what you want (they are not really used anyways).

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Thanks, I wasn't aware of this discussion. All the other problems I had to solve regarding contour integration were considerably easier so far, so I guess I expected a short solution - but a more abstract formula is much nicer, of course. –  Eli Jul 9 '11 at 19:10
    
This is why I wrote this into the community wiki, there might be a shortcut to get your result but I doubt it. Basically the exponents $m$ and $n$ are not essential to the calculation as you can see in the problems i linked in the question, they are solved in a similar way. –  Listing Jul 9 '11 at 19:17

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