How can one show that \begin{equation*} \int_0^{\infty} \frac{\cos (xt)}{1+t^2} dt = \frac{\pi}{2}e^{-x} \end{equation*} using Laplace Transform?
I tried to use the formula $$\int_0^{\infty} \frac{f(x)}{x} dx = \int_0^{\infty} F(s) ds$$ where $F$ denotes the Laplace Transform of $f$ but to no avails.
Any help would be appreciated.
On
Not $\mathcal F(x) =\mathcal L(ix)-\mathcal L(-ix)$ hence we use Fourier transform in place of Laplace transform,
But we have, \begin{split} \mathcal F(e^{-|t|})(x) = \int_{-\infty}^{\infty}e^{-|t|}e^{-ix t}\,dt &=&\int_{-\infty}^{0}e^{t}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-t}e^{-ix t}\,dt\\ &=&\left[ \frac{e^{(1-ix)t}}{1-ix} \right]_{-\infty}^0-\left[\frac{e^{-(1+ix)t}}{1+ix} \right]_{0}^{\infty}\\ &=&\frac{1}{1-ix}+\frac{1}{1+ix}\\ &=&\frac{2}{x^2+1}. \end{split} Then, $$ \begin{align} e^{-|a|}=\mathcal F^{-1}\left( \frac{2}{x^2+1}\right)(a) &=\frac{1}{2\pi}\int_\Bbb R \frac{2}{x^2+1}e^{ix a}\,dx = \frac{1}{\pi}\int_\Bbb R\frac{e^{ix a}}{x^2+1}\,dx \\&=\frac{1}{\pi}\int_\Bbb R\frac{\cos a x}{x^2+1}\,dx = \frac{2}{\pi}\int_0^\infty\frac{\cos ax}{x^2+1}\,dx \end{align} $$ Given that, as $x\mapsto\sin ax $ is an old function we have, $$\int_\Bbb R \frac{\sin{a x}}{x^2+1}dx= 0.$$
Thus we have, $$ \int_0^\infty\frac{\cos ax}{x^2+1}\,dx =\frac{\pi}{2}e^{-|a|} $$
Recall that, if we consider the Fourier transform $$\mathcal Ff (a) =\int_\Bbb R e^{-ia x}f(x)dx$$ then its Fourier inverse is defined as $$\mathcal F^{-1}f (x) =\frac{1}{2\pi}\int_\Bbb R e^{it x}f(t)dt.$$
Substituting $\cos xt=\frac{e^{ixt}+e^{-ixt}}{2}$ we have:$$I=\int_0^{\infty} \frac{\cos (xt)}{1+t^2} dt=\frac{1}{2}\int_0^{\infty} \frac{e^{ixt}+e^{-ixt}}{1+t^2} dt$$Denote by $H(x)$ the Laplace transform of $h(t)=\frac{1}{1+t^2}u(t)$. Therefore $I=\frac{1}{2}(H(x)+H(-x))$ which is the Laplace transform of $\frac{1}{2}(h(t)+h(-t))=\frac{0.5}{1+t^2}$. Therefore $I$ is the Laplace transform of $\frac{0.5}{1+t^2}$ and we have $I=\frac{1}{4}2\pi e^{-|x|}$ or for $x\ge 0$ , $I=\frac{\pi}{2}e^{-x}$