Evaluate $\int\limits_0^{\infty} \frac{t^2dt}{(t^2+1)^2}$

Question

Here's how I proceeded:

Set $f(z) := \frac{z^2}{(z^2+1)^2}$, $g(t):=\frac{t^2}{(t^2+1)^2}$.

$(z^2+1)^2=[(z+i)(z-i)]^2$, so $f(z)$ has poles of order $2$, and the singular points are $i$ and $-i$. Now one needs to find the residue at $i$ in order to evaluate this integral over the contour of radius $R$, located in the upper half of the complex plane, as well as on the real line from $R$ to $-R$.

The question is: how to find the residue at $i$?

$Res_{z=i}[f(z)]=\lim\limits_{z\to i} \frac{d}{dz}(z-i)^2 \frac{z^2}{[(z-i)(z+i)]^2}=2\lim\limits_{z\to i}(z-i) \frac{z^2}{[(z-i)(z+i)]^2}=2\lim\limits_{z\to i} \frac{z^2}{(z-i)(z+i)^2} $

But here's where I get stuck. Would appreciate some clarification.

Answer 1

You made a mistake at the derivative. Note that $$\left(\frac{z^2}{(z+i)^2}\right)' = \frac{2z(z+i)^2-2z^2(z+i)}{(z+i)^4}=\frac{2zi}{(z+i)^3}.$$ Hence the residue you want is $\frac{1}{4i}$.

Answer 2

I've made the same error once, because I applied the theorem wrong.

The theorem is as follows:

Let $z_0$ be a pole of order $N$ of a function $f$, then $$\mathop{\text{res}}_{z=z_0} f(z) = \frac{1}{(N-1)!}\cdot \lim_{z\to z_0}\left((z-z_0)^N f(z) \right)^{(N-1)}$$

Notice how the derivative is taken on the function $(z-z_0)^N f(z)$, the $(z-z_0)^N$ multiplication takes care of the singularity.

Btw, Wolfram supports residue calculation:

residue of z^2/(z^2+1)^2 at z=i

Results in $\frac{-i}{4}$

Answer 3

By integration by parts $$ \int_{0}^{+\infty}\frac{t^2}{(t^2+1)^2}\,dt = \int_{0}^{+\infty}\frac{t}{2}\cdot\frac{2t}{(t^2+1)^2}\,dt = \frac{1}{2}\int_{0}^{+\infty}\frac{dt}{1+t^2}=\color{red}{\frac{\pi}{4}},$$ hence the problem is quite trivial, with or without complex analytic techniques.