# Determine the integral $\int_0^\infty \frac{\mathrm{d}x}{(x^2+1)^2}$ using residues.

Determine the integral

$$\int_0^\infty \frac{\mathrm{d}x}{(x^2+1)^2}$$

using residues. This is from Section 79, Brown and Churchill's Complex Variables and Applications.

In order to do this. We should first consider the complex analogue of this function $f(z) = \frac{1}{(z^2+1)^2}$. We see then that there are two singularities at $z = i$ and at $z = -i$.

In order to evaluate this integral we should consider a line which lies on the real axis and an enclosing contour creating an simple closed curve. Assuming this contour lies on the upper half plane, it encloses only one singular point existing at $z = i$ therefore we need only calculate the residue at this point.

This is as far as I've gotten. I don't know how to calculate this residue. It seems that every method of tried short of a brute force expansion using Laruent's series has failed. For example:

$$\mathrm{Res}_{z = i} \frac{1}{(z^2+1)^2} = \frac{\phi(i)}{z-i}$$

evaluating $\phi(i)$ gives complex infinity. Useless for residue calculation.

How do I get this residue?

• ...and what in the world is $\,\phi\,$ , anyway? Commented Apr 30, 2013 at 22:56
• When you used brute force to calculate the Laurent series, what terms did you get? Commented Apr 30, 2013 at 22:59

Hint I always think of these in terms of partial fractions. $$f(z)=\frac{1}{(z^2+1)^2} = \frac{1}{(z+i)^2 (z-i)^2} = \frac{A}{(z-i)} + \frac{B}{(z-i)^2}+\frac{C}{(z+i)} + \frac{D}{(z+i)^2}$$

Can you find $(A,B,C,D)$ that make that true? Then do the residues pop out at you?

Edit

To continue $$B = \left.f(z)(z-i)^2\right|_{z=i} = \left.\frac{1}{(z+i)^2}\right|_{z=i} = -\frac{1}{4}$$ $$D = \left.f(z)(z+i)^2\right|_{z=-i} = \left.\frac{1}{(z-i)^2}\right|_{z=-i} = -\frac{1}{4}$$

Now that you know those two, move them over to the other side and see what's left: $$g(z)=f(z)+\frac{1}{4(z-i)^2}+\frac{1}{4(z+i)^2}= \frac{A}{(z-i)} + \frac{C}{(z+i)}$$ $$g(z)=\frac{1}{2 (z-i) (z+i)}=\frac{A}{(z-i)} + \frac{C}{(z+i)}$$ And finally solve for $A$ and $C$ the same way as before: $$A = \left.g(z)(z-i)\right|_{z=i} = \left.\frac{1}{2(z+i)}\right|_{z=i} = \frac{1}{4i}$$ $$C = \left.g(z)(z+i)\right|_{z=-i} = \left.\frac{1}{2(z-i)}\right|_{z=-i} = -\frac{1}{4i}$$

Giving a final expression: $$f(z)= \frac{1}{4i(z-i)} - \frac{1}{4(z-i)^2}-\frac{1}{4i(z+i)} - \frac{1}{4(z+i)^2}$$

Now the residues should jump out at you. As a final note, formally, the expressions above in which poles are cancelled by multiplication and then things are evaluated at the former poles should have a limit in them, but this is shorthand when working this kind of problem.

Edit Corrected a sign error in the $C$ coefficient. The partial fraction expansion checks out now.

Since the poles of the function are double ones you have to evaluate, for example:

$$\lim_{z\to i}\;\left((z-i)^2\frac1{(z^2+1)^2}\right)'=\lim_{z\to i}\left(\frac1{(z+i)^2}\right)'=\lim_{z\to i}\;-\frac2{(z+i)^3}=\frac1{4 i}\ldots$$