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

complex integrationcomplex-analysis

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?

Best Answer

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.

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