I have to calculate $\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)^3}$ using the residue theorem. Doing it step by step I get:

  1. Let $f(z) = \frac{1}{(z^2 + 1)^3}$. I know I can use the contour $C = C_1 + C_R$ where $C_1$ is a straight line on the real axis from $-R$ to $R$ and $C_R$ is a semicircle from $-R$ to $R$. So I get $\int_{C_1 + C_R} f(z)dz = 2\pi i \sum Res(f,z_0)$.

  2. The integral over the contour $C_R$ is for $R \to \infty$ equal to $0$. So we are left with the integral over $C_1$. We have then $\lim_{R \to \infty} \int_{C_1} f(z)dz = \lim_{R \to \infty} \int_{-R}^{R}f(x)dx = \int_{-\infty}^{\infty}f(x)dx = 2\pi i \sum Res(f,z_0)$

  3. I found that the residues can be determined by the following equation: $$Res(f,z_0) = \frac{1}{(N-1)!}\lim_{z\to z_0}(\frac{d}{dz})^{N-1} (z-z_0)^{N}f(z)$$ where $N$ is the order of the poles. Since I have $f(z) = \frac{1}{(z^2 + 1)^2}$, the poles are at $z_0 = \pm i$ but in my case I need only the $z_0 = i$. The order is $2$. Putting this in the equation and rewriting $z^2 + 1 = (z-1)(z+i)$, I am left with $$Res(f,i) = \lim_{z\to i}\frac{d}{dz} (z-i)^{2}\frac{1}{(z -i )^3(z +i )^3} = \lim_{z\to i}\frac{d}{dz} \frac{1}{(z - i )(z +i )^3}$$.

Here I'm stuck, because after I determine the derivative I get $$Res(f,i) = \lim_{z\to i} -\frac{1}{(z - i )^2(z +i )^3} - \frac{3}{(z - i )(z +i )^4}$$ and if I put $z = i$ in the equation, I get $-\frac{1}{0} - \frac{3}{0}$. Where did I make a mistake?


1 Answer 1


Your mistake was the order of the pole $z_0 = i$. It should be $3$ instead of $2$.

The residue formula signals you did something wrong when the product $(z - z_0)^N f(z)$ still has pole terms.

  • $\begingroup$ thanks for the reply! so this is basically my check if I got the pole number right? If the product still has pole terms like you wrote it? And how to determine the pole number correctly? I assuemd it is $2$ because in $f(z)$ I have the expresion $z^2$ $\endgroup$
    – syphracos
    Nov 10, 2022 at 11:11
  • $\begingroup$ @syphracos I guess you should go back to definitions: order of a pole $z_0$ of $f(z)$ is defined by the Laurent expansion of $f(z)$ around the disc punctured at $z_0$, assuming $f(z)$ is holomorphic in this disc of course (o.w. it is an essential singularity). $\endgroup$
    – saru
    Nov 10, 2022 at 11:24
  • $\begingroup$ @syphracos If we put some more effort, we get an even easier criterion: $z_0$ is a pole of order $N$ iff $\lim_{z \to z_0} (z - z_0)^N f(z)$ exists and is non-zero. $\endgroup$
    – saru
    Nov 10, 2022 at 11:28
  • $\begingroup$ @syphracos this criterion above is also why you can immediately see you are doing something wrong when computing the residue formula. When you use a value < pole order, you get a nonexisting limit (which is what you get initially with the division by zero). $\endgroup$
    – saru
    Nov 10, 2022 at 11:34
  • $\begingroup$ okay I see what you mean. Thank you! $\endgroup$
    – syphracos
    Nov 10, 2022 at 11:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .