compute $\int_0^{\infty}\frac{\cos x}{x^2+a^2}dx$ in complex plane without using the residue compute
$\int_0^{\infty}\frac{\cos x}{x^2+a^2}dx$
in complex plane without using the residue
I know that one way is to calculate this integral
$\int_{c(r)}\dfrac{e^{iz}}{z^2+a^2}$
over half a circle, ( assume a≠0 here). At the end use the fact that $\cos x$ is the real part of $e^{ix}$
i've come to this
$\int_{c(r)}\dfrac{e^{iz}}{z^2+a^2}=\int_{-R}^R\dfrac{e^{iz}}{z^2+a^2}+\int_{\lambda(r)}\dfrac{e^{iz}}{z^2+a^2} $
But at this point they use the ressidue,
Is there a way to calculate it without using the residue?
I know that Jordan's lema can be useful
$\lim_{R\longrightarrow \infty}\int_{C(R)}f(z)e^{iz}dz=0$
 A: $$\frac{\cos (x)}{x^2+a^2}=\frac{\cos (x)}{(x+ia)(x-ia)}=\frac i{2a}\Bigg[\frac{\cos (x)}{x+i a}-\frac{\cos (x)}{x-i a} \Bigg]$$
$$I_k=\int \frac{\cos (x)}{x+k}\,dx$$ Make $x+k=t$ and expand the cosine
$$I_k=\sin (k)\int\frac{ \sin (t)}{t}dt+\cos (k)\int\frac{ \cos (t)}{t}dt$$
$$I_k=\sin (k)\, \text{Si}(t)+ \cos (k)\,\text{Ci}(t)+C$$ Then
$$\int\frac{\cos (x)}{x^2+a^2}\,dx=\frac{\sinh (a) (\text{Si}(i a-x)-\text{Si}(i a+x))-i \cosh (a) (\text{Ci}(x-i
   a)-\text{Ci}(i a+x))}{2 a}$$
$$\int_0^\infty\frac{\cos (x)}{x^2+a^2}\,dx=\frac{\pi  }{2 a}e^{-a}$$
A: Well, this is my first answer. I hope my answer will help you.
Instead of using the residue theorem, we can use different methods to calculate this integral which is known as Laplace integral.
Laplace transform:
Let $I(b) = \dfrac{\cos bx}{a^2 + x^2}$,
$$
\begin{align}
\mathcal{L}[I(b)] &= \int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos bx}{a^2+x^2}e^{-pb}\ \mathrm{d}x\mathrm{d}b\\
& =\int_{0}^{\infty}\cos bx e^{-pb}\ \mathrm{d}b \int_{0}^{\infty}\frac{1}{a^2+x^2} \ \mathrm{d}x\\
&=\int_{0}^{\infty}\frac{p}{p^2+x^2} \frac{1}{a^2+x^2} \ \mathrm{d}x\\
&=\frac{\pi}{2a}\frac{1}{a+p}
\end{align}
$$
Note that $\mathcal{L}[e^{-ax}] = \dfrac{1}{a+p}$, so take the inverse Laplace transform of both sides,
$$
\boxed{
I(b) = \int_{0}^{\infty}\frac{\cos bx}{a^2+x^2} = \frac{\pi}{2a}e^{-ab}}
$$
Fourier transform:
Fourier cosine transform:
$$
F(\xi )= \sqrt{\frac{2}{\pi}} \int_{0}^{\infty}f(x)\cos \xi x\ \mathrm{d}x\\
f(x) = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty}F(\xi) \cos\xi x \ \mathrm{d}\xi
$$
Let $f(x) = e^{-ax}$,
$$
F(\xi )= \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} e^{-ax} \cos \xi x\ \mathrm{d}x = \sqrt{\frac{2}{\pi}}\frac{a}{a^2+\xi^2}
$$
Take the inverse transformation, then
$$
e^{-ax} = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty}\sqrt{\frac{2}{\pi}} \frac{a}{a^2+\xi^2} \cos \xi x \ \mathrm{d}x\\
\Rightarrow \boxed{\int_{0}^{\infty}\frac{\cos bx}{a^2+x^2} = \frac{\pi}{2a}e^{-ab}}
$$
Or you can use the Parseval's theorem for Fourier transform. But it seems that it also needs to use the residue theorem.
