# \int_{0}^\infty {\cos(xt)\over 1+t^2}dt I'm supposed to solve this using Laplace Transformations.

Jazmine Sweeney 2022-04-30 Answered
${\int }_{0}^{\mathrm{\infty }}\left\{\mathrm{cos}\left(xt\right)over1+{t}^{2}\right\}dt$
I'm supposed to solve this using Laplace Transformations.
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Let f(x) denote the integral, and assume temporarily that x>0. This makes no difference since f(x) is even by definition. Then its Laplace transform Lf(s) defines a continuous function on $s\in \left(0,\mathrm{\infty }\right)$ (in fact, it defines an analytic function on R(s)>0). Thus we may assume further that $s\ne 1$ and then rely on the continuity argument. Then
$\mathcal{L}f\left(s\right)={\int }_{0}^{\mathrm{\infty }}f\left(x\right){e}^{-sx}dx$
$={\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(xt\right)}{1+{t}^{2}}{e}^{-sx}dtdx$
$\stackrel{\cdot }{=}{\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(xt\right)}{1+{t}^{2}}{e}^{-sx}dxdt$
$={\int }_{0}^{\mathrm{\infty }}\frac{1}{1+{t}^{2}}\frac{s}{{s}^{2}+{t}^{2}}dt$
$=\frac{s}{1-{s}^{2}}{\int }_{0}^{\mathrm{\infty }}\left(\frac{1}{{s}^{2}+{t}^{2}}-\frac{1}{1+{t}^{2}}\right)dt$
$=\frac{s}{1-{s}^{2}}\frac{\pi }{2}\left(\frac{1}{s}-1\right)=\frac{\pi }{2}\frac{1}{1+s}$
Here, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for $s\ne 1$, it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find that
$f\left(x\right)=\frac{\pi }{2}{e}^{-x}$