# Evaluating $\int_{-\infty}^{\infty}\frac{1}{(x^2+b^2)^2}dx$

Find $$\int_{-\infty}^{\infty}\frac{1}{(x^2+b^2)^2}dx$$

We see that the only poles are at $x=\pm bi$. Integrating over the semicircular contour implies that it is equal to $2\pi i*Res_{(+bi)}$ because the integral over the circular arc is $0$. So now I just need to calculate the residue at $+bi$. Can anyone give hints on how to do this? I was thinking about multiplying by $(x-bi)^2$ and then evaluating at $bi$ but I'm not sure why this even makes sense.

• How do you usually find residues? Nov 24, 2014 at 13:04
• We can find the $a_{-1}$ term of the laurent expansion. Would this be too messy in this case though? I'm also not sure how to find the laurent expansion in this case. Nov 24, 2014 at 13:05
• $x=b\tan \theta \implies dx=b\sec^2 \theta d\theta \implies \therefore\displaystyle\int\frac{dx}{(x^2+b^2)^2}=\dfrac{1}{b^4}\displaystyle \int\cos^2 \theta d\theta$
– user170039
Nov 24, 2014 at 13:06
• @BobbyJones If $z_0$ is a pole of order $k$ of $\varphi$, then $$\text{Res}(\varphi, z_0)=\dfrac 1{(k-1)!}\lim \limits_{z\to z_0}\left[\dfrac{\mathrm d^{k-1}}{\mathrm dz^{k-1}}\left(z\mapsto (z-z_0)^k\varphi(z)\right)(z)\right].$$ See an explanation in the first part of this answer. Nov 24, 2014 at 13:07
• so it would just be $\lim_{z\to bi}\frac{d}{dz}(\frac{1}{(z+bi)^2})(z)=1/4b^2$. Assuming I did the calculations correctly Nov 24, 2014 at 13:22

A possible contour is the semi circle in the upper half plane with a semi circle around the origin. Let $\int_{\Gamma}$ be large semi circle and $\int_{\gamma}$ be the small semi circle. Let $R$ be the radius of the $\Gamma$ and $\epsilon$ the radius of $\gamma$. Now as $R\to\infty$, $\int_{\Gamma}\to 0$ and similarly as $\epsilon\to 0$, $\int_{\gamma}\to 0$ by the estimation lemma. Take the orientation of the contour to be counter clockwise. Therefore, we can write you integral as \begin{align} \int_{-\infty}^{\infty}f(x)dx &= \int_{-\infty}^{\infty}f(z)dz\\ &= \int_{\Gamma}+\int_{\gamma} + \int_{-\infty}^{\infty}f(z)dz\\ &= 2\pi i\sum_{\text{UHP}}\text{Res}\\ \int_{-\infty}^{\infty}f(z)dz &=2\pi i\sum_{\text{UHP}}\text{Res} \end{align} As you have been told in the comments, the residue for a pole which isn't simple is $$2\pi i\lim_{z\to z_0}\frac{1}{(k - 1)!}\frac{d^{k-1}}{dz^{k-1}}(z - z_0)^kf(z)$$ In your case, the pole of order two in the upper half plane is $z_0 = ib$. That is, \begin{align} 2\pi i\lim_{z\to z_0}\frac{1}{(k - 1)!}\frac{d^{k-1}}{dz^{k-1}}(z - z_0)^kf(z) &= 2\pi i\lim_{z\to ib}\frac{1}{1!}\frac{d}{dz}\frac{1}{(z + ib)^2}\\ &= 2\pi i\lim_{z\to ib}\frac{-2}{(z+ib)^3} \end{align}

• The $2\pi i$ factor isn't part of the residue. Nov 24, 2014 at 14:24
• @GitGud it is part of Cauchy integral formula. Since we have no poles on the real axis which we are only taking a semi circle around, we pick up $2\pi i$ instead of the $\pi i$. Certain books will define the residue enclosed in a contour in this manner and have a separate definition for poles on the real axis which are then defined as $\pi i$ sum residues. Nov 24, 2014 at 14:26