Need help evaluating this specific definite integral

Question

I am trying to show that:

$$\int_{0}^{\infty} \frac{x^2}{(x^2 + a^2)^2} dx = \frac{\pi}{4a}$$

I know that this can be represented as $f(z) = \frac{z^2}{(z^2 + a^2)^2} dz$, and that this function will have poles at $z = \pm ia$. And I also know that this integral will equal the sum of the residues at the two poles multiplied by $2\pi i$, $z = ia$ and $z = -ia$. But from here I am a little confused as to how to get this answer of $\frac{\pi}{4a}$.

When I calculate the residue at $z = ia$ with order $= 2$, I get:

$$Res(f(z); z_0=ia) = \frac{1}{(2-1)!} \lim_{z \to ia} (z-ia)^2 \frac{d}{dz}\left( \frac{1}{(z^2 + a^2)^2} \right),$$ which equals:

$$\lim_{z \to ia} \frac{-(z-ia)^2}{4z(z^2 + a^2)^3}$$

But this evaluates to zero in the denominator when taking the limit. I am also getting zero in the denominator for the pole at $z = -ia$. Can anyone see where I am going wrong and help me fix this?

Answer 1

I believe that I have found the way to do it in spirit of how my course is taught. While @dan_fulea answer is correct, we haven't been taught to evaluate integrals using O(h) functions. This is how I I ended up solving this:

Prove that: $\int_{0}^{\infty} \frac{x^2}{(x^2 + a^2)^2} = \frac{\pi}{4a}$

We can solve this by the fact that:

$\int_{0}^{\infty} \frac{x^2}{(x^2 + a^2)^2} = \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^2}{(x^2 + a^2)^2} --- Eq (1)$

First we want to find if there are any singularities/poles to this function, so let it become a function of z. $f(z) = \frac{z^2}{(z^2 + a^2)^2}$ , we will find that the roots are $z = +/- ia$ both of Order = 2. From there we can calculate this integral by calculating the sum of the residues of the singularities/poles. We will only consider the pole $z = + ia$, as the contour we are integrating over will be a semi circle only covering the upper half plane, and therefore will only enclose the pole $z = + ia$. We will call this integral I, and the pole's order of 2 will be equal to n:

$I = 2\pi i (Res(f(z);z_0 = ia))$ $= 2\pi i (\frac{1}{(n-1)!})\lim_{y\to z_o} \frac{d^{n-1}}{dz^{n-1}}[(z-z_0)^n (\frac{z^2}{(z^2 + a^2)^2})]$

$= 2\pi i (\frac{1}{(2-1)!})\lim_{y\to ia} \frac{d}{dz}[(z-ia)^2 (\frac{z^2}{z^2 + a^2})]$

Now to help simplify this, we can use the fact that $(z^2 + a^2) = (z+ia)(z-ia)$. Substituting into our equation we get:

$= 2\pi i \lim_{y\to ia} \frac{d}{dz}[(z-ia)^2 (\frac{z^2}{(z+ia)^2(z-ia)^2})]$ . Now cancel out the $(z-ia)^2$ terms to get:

$= 2\pi i \lim_{z \to ia} \frac{d}{dz}(\frac{z^2}{(z + ia)^2})$ . Taking the derivative we will get:

$= 2\pi i \lim_{z \to ia}(\frac{2iaz}{(z + ia)^3})$

$= 2 \pi i \frac{2ia(ia)}{(ia + ia)^3}$

$ = 2 \pi i \frac{2i^2a^2}{(2ia)^3}$

$ = 2 \pi i \frac{2i^2a^2}{8i^3a^3}$

$= \frac{4 \pi i}{8ia}$

$= \frac{2\pi}{4a}$

Now divide by two to satisfy Eq (1):

$= \frac{2\pi}{8a} = \frac{\pi}{4a}$

Thank you to all who took their time to try to help me solve this!

Answer 2

The function $f$ is a true function, it does not have any $dz$ in it: $$ f(z) =\frac {z^2}{(z^2+a^2)^2}\ , $$ and to calculate its residue in $ia$, let us write explicitly $z=ia+h$, then get a Laurent power series in $h$. I will type the computation explicitly, using small steps. $$ \begin{aligned} f(ia+h) &= \frac {(ia+h)^2}{(a^2+(ia+h)^2)^2} =\frac{-a^2+2iah+h^2}{(2iah+h^2)^2} \\ &= \frac1{(2ia)^2}\cdot\frac 1{h^2}\cdot (-a^2+2iah+O(h^2))\cdot\frac 1{\left(1+\frac h{2ia}\right)^2} \\ &= \frac1{-4a^2}\cdot\frac 1{h^2}\cdot (-a^2+2iah+O(h^2))\cdot\left(1-\frac h{2ia}+O(h^2)\right)^2 \\ &= \frac14 \cdot\frac 1{h^2}\cdot \left(1-\frac {2i}a h+O(h^2)\right)\cdot\left(1+\frac ia h+O(h^2)\right) \\ &= \frac14 \cdot\frac 1{h^2}\cdot \left(1\bbox[yellow]{\ -\frac ia h\ }+O(h^2)\right) \ , \end{aligned} $$ and the residue comes from the yellow term: $$ \operatorname{Res}_{z=ia}f(z)=-\frac i{4a}\ . $$ This leads to $$ \int_0^\infty f(x)\; dx = \lim_R \int_0^R f(x)\; dx = \frac 12\lim_R \int_{-R}^R f(x)\; dx \overset!= \frac 12\lim_R \int_{[-R,R]\cup \frown_R} f(z)\; dz \\ =\frac 12\lim_R 2\pi i\operatorname{Res}_{z=ia}f(z) = \frac 12\cdot2\pi i\cdot\frac{-i}{4a} =\frac \pi{4a}\ . $$