# calculate the integral $\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)^3}$ by using the residue theorem

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?

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.

• 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 $z^2$ Nov 10, 2022 at 11:11
• @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).
– saru
Nov 10, 2022 at 11:24
• @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.
– saru
Nov 10, 2022 at 11:28
• @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).
– saru
Nov 10, 2022 at 11:34
• okay I see what you mean. Thank you! Nov 10, 2022 at 11:36