compute $\int_0^{\infty}\frac{\cos x}{x^2+a^2}dx$ in complex plane without using the residue

Question

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

$\lim_{R\longrightarrow \infty}\int_{C(R)}f(z)e^{iz}dz=0$

Answer 1

$$\frac{\cos (x)}{x^2+a^2}=\frac{\cos (x)}{(x+ia)(x-ia)}=\frac i{2a}\Bigg[\frac{\cos (x)}{x+i a}-\frac{\cos (x)}{x-i a} \Bigg]$$

$$I_k=\int \frac{\cos (x)}{x+k}\,dx$$ 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 $$\int\frac{\cos (x)}{x^2+a^2}\,dx=\frac{\sinh (a) (\text{Si}(i a-x)-\text{Si}(i a+x))-i \cosh (a) (\text{Ci}(x-i a)-\text{Ci}(i a+x))}{2 a}$$ $$\int_0^\infty\frac{\cos (x)}{x^2+a^2}\,dx=\frac{\pi }{2 a}e^{-a}$$

Answer 2

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.