Improper integral using residue

[player1_1_1]

Homework Statement

integral: $$\int\limits_0^\infty\frac{\mbox{d}x}{\left(x^2+1\right)\left(x^2+4\right)}$$

The Attempt at a Solution

normally i would do $$I=\frac12\int\limits_{-\infty}^\infty\frac{\mbox{d}x}{\left(x^2+1\right)\left(x^2+4\right)}$$ and now count residues but is there any other thing what i can do without making it in $$x\in[-\infty,\infty]$$? what if i had $$\int\limits_a^\infty\frac{\mbox{d}x}{\left(x^2+1\right)\left(x^2+4\right)}$$ where $$a$$ is real, positive number?

 

[snipez90]

I think this is simple enough that you can actually compute the antiderivative (use partial fractions). As for complex integration techniques, I think there is no convenient way to work out that last integral. The reason you need the symmetric real interval is to shift the contour and close it off in order to apply the residue theorem. For the original integral, you know the relevant contour is simply a symmetric line segment running from -R to R (R > 0) and a semicircle of corresponding radius attached either above or below the line segment, with appropriate orientation. The most naive way to try to tackle the last integral might be to consider a quarter circle with vertical and horizontal line segments (the horizontal part sitting on the real line; the vertical part along x = a), but the problem is that the integral over the vertical segment won't vanish. Unfortunately I don't really have a better answer right now.

 

[HallsofIvy]

You could take your contour up the positive real axis from 0 to the real number R, then around the quarter circle |z|= R to Ri, then down the imaginary axis to 0. Finally, of course, take the limit as R goes to infinity. The integral around the quarter circle should go to 0 and the integral down the imaginary axis will be related to the integral on the positive real axis.

 

[snipez90]

You can't have the contour running down the imaginary axis since the poles are on the imaginary axis. But translate that contour to the right by a > 0 and you have what I described before. I still don't think this is a good approach because well practically speaking, you're not calculating residues explicitly anymore (this contour does not enclose any singularities) so you're essentially applying Cauchy's theorem. The real difficulty I think is relating the integral over the vertical segment to the integral over the horizontal segment and ensuring that you do actually get the intended value of the integral out of it.