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

Question

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?

Answer 1

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 $\frac{1}{4} \pi (a+1) e^{-a}$.

Answer 2

Alternative Method

First of all, we let $t>0$, then $$ \begin{aligned} I(a) & =\int_{-\infty}^{\infty} \frac{\cos (t x)}{a+x^2} d x \text {, where }a>0 \\ & =\Re \int_{-\infty}^{\infty} \frac{e^{t x i}}{a+x^2} d x \\ & =\Re\left[2 \pi i\left(\lim _{z \rightarrow \sqrt{a }i} \frac{e^{t z i}}{2 z}\right)\right] \\ & =\frac{\pi}{\sqrt{a}} e^{-\sqrt{a} t} \end{aligned} $$ Differentiating both sides w.r.t. $a$ yields $$ \begin{aligned} \int_{-\infty}^{\infty} \frac{\cos (t x)}{\left(1+x^2\right)^2} d x &= -\left.\frac{d}{d a}(I(a))\right|_{a=1} \\ & =-\left.\frac{d}{d a}\left(\frac{\pi}{\sqrt{a}} e^{-\sqrt{a} t}\right)\right|_{a=1} \\ & =\left.\frac{\pi e^{-t \sqrt{a}}(t \sqrt{a}+1)}{2 a^{\frac{3}{2}}}\right|_{a=1} \\ & =\frac{\pi}{2} e^{-t}(t+1) \end{aligned} $$ For $t<0$, replacing $t$ by $-t$ gives $$ \int_{-\infty}^{\infty} \frac{\cos (t x)}{\left(1+x^2\right)^2} d x =\frac{\pi}{2} e^t(-t+1) $$ Conclusively, we have $$ \boxed{\int_{-\infty}^{\infty} \frac{\cos (t x)}{\left(1+x^2\right)^2} d x =\frac{\pi}{2} e^{-|t|}(|t|+1)} $$ In general, differentiating both sides w.r.t. $a$ by $(n-1)$-times yields $$\begin{aligned} \int_{-\infty}^{\infty} \frac{\cos (t x)}{\left(a+x^2\right)^n} & =\left.\frac{(-1)^{n-1}}{(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\pi}{\sqrt{a}} e^{-\sqrt{a} t}\right)\right|_{a=1} \\ & = \boxed{\left.\frac{2(-1)^n \pi}{t(n-1) !} \frac{d^n}{d a^n}\left(e^{-\sqrt{a} t}\right)\right|_{a=1}} \end{aligned} $$ where $t>0$. In particular, $$ \int_{-\infty}^{\infty} \frac{\cos (t x)}{\left(1+x^2\right)^3} d x=\frac{e^{-|t|}\left(t^2+3 |t|+3\right) \pi}{8} $$