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.

  • 1
    $\begingroup$ How do you usually find residues? $\endgroup$
    – Git Gud
    Nov 24, 2014 at 13:04
  • $\begingroup$ 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. $\endgroup$ Nov 24, 2014 at 13:05
  • $\begingroup$ $x=b\tan \theta \implies dx=b\sec^2 \theta d\theta \implies \therefore\displaystyle\int\frac{dx}{(x^2+b^2)^2}=\dfrac{1}{b^4}\displaystyle \int\cos^2 \theta d\theta$ $\endgroup$
    – user170039
    Nov 24, 2014 at 13:06
  • 1
    $\begingroup$ @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. $\endgroup$
    – Git Gud
    Nov 24, 2014 at 13:07
  • $\begingroup$ 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 $\endgroup$ Nov 24, 2014 at 13:22

1 Answer 1


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}

  • $\begingroup$ The $2\pi i$ factor isn't part of the residue. $\endgroup$
    – Git Gud
    Nov 24, 2014 at 14:24
  • $\begingroup$ @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. $\endgroup$
    – dustin
    Nov 24, 2014 at 14:26

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