# $\int_{0}^{\infty}{\dfrac{\cos(ax)}{(x^2 + 1)^2}dx}$

I have a contour integral problem I need to solve, but I don't know the answer, so I wanted to verify that my work is correct.

$$\int_{0}^{\infty}{\frac{\cos(ax)}{(x^2 + 1)^2}dx}$$

For this one, the function being integrated is even, so I can just take the integral over the entire real line and multiply by $\dfrac{1}{2}$. That is $\int_{0}^{\infty}{\dfrac{\cos(ax)}{(x^2 + 1)^2}dx} = \dfrac{1}{2}\int_{-\infty}^{\infty}{\dfrac{\cos(ax)}{(x^2 + 1)^2}dx}$. In the upper half-plane, the function being integrated has a double pole at $i$. Therefore, I want to say that this is true:

$$\int_{-\infty}^{\infty}{\frac{\cos(ax)}{(x^2 + 1)^2}dx} = \operatorname{Re} [2 \pi i\ \operatorname{Res}(\dfrac{e^{iz}}{(x^2 + 1)^2}, i)]$$

My solution yields: $\int_{0}^{\infty}{\dfrac{\cos(ax)}{(x^2 + 1)^2}dx} = \dfrac{\pi}{4e}$

I have no way to verify the correctness of my answer, so is this correct or have I made a mistake somewhere?

• You loose and expotent term. $\frac{1}{4} \pi e^{-\left| a\right| } (\left| a\right| +1)$ Remember that you have a double pole. – Caran-d'Ache May 15 '13 at 9:36

You have: $$\int_{-\infty}^{\infty}{\frac{\cos(ax)}{(x^2 + 1)^2}dx} = \operatorname{Re} \left[2 \pi i\ \operatorname{Res}\left(\dfrac{e^{i a z}}{(z^2 + 1)^2}, i\right)\right]$$ And $$\operatorname{Res}\left(\dfrac{e^{i a z}}{(z^2 + 1)^2}, i\right)=\operatorname{Res}\left(\dfrac{e^{i a z}}{(z +i)^2(z -i)^2}, i\right)=\lim_{z \to i} \frac{d}{dz}\left( (z-i)^{2}f(z) \right)$$ where $f(z)=\dfrac{e^{i a z}}{(z +i)^2(z -i)^2}$.
So $$\frac{d}{dz}\left( (z-i)^{2}\dfrac{e^{i a z}}{(z +i)^2(z -i)^2} \right)=\frac{i a e^{i a z}}{(z+i)^2}-\frac{2 e^{i a z}}{(z+i)^3}$$ Taking the limit, multiplying by $2\pi i$, taking the real part and finally multiplying by $\frac{1}{2}$ will give you $\frac{1}{4} \pi (a+1) e^{-a}$.