# Opt1

Find the associated power spectrum given the covariance function ${\displaystyle B_{x}(\tau )=\exp \left(-\left|{\frac {\tau }{\tau _{0}}}\right|\right),\tau _{0}>0\,}$.

The power spectrum is defined as ${\displaystyle S_{x}(\omega )={\frac {1}{\pi }}\int _{0}^{\infty }B_{x}(\tau )\cos(\omega \tau )\,d\tau \,}$.

${\displaystyle ={\frac {1}{\pi }}\int _{0}^{\infty }\exp \left(-\left|{\frac {\tau }{\tau _{0}}}\right|\right)\cos(\omega \tau )\,d\tau \,}$

${\displaystyle ={\frac {1}{\pi }}\int _{0}^{\infty }\exp \left(-{\frac {\tau }{\tau _{0}}}\right)\cos(\omega \tau )\,d\tau \,}$ since ${\displaystyle \tau _{0}>0\,}$ and ${\displaystyle \tau >0\,}$.

A useful integral relation is:

${\displaystyle \int _{0}^{\infty }x^{\mu -1}e^{-ax}\cos(bx)\,dx={\frac {\Gamma (\mu )}{(a^{2}+b^{2})^{\mu /2}}}\cos(\mu \arctan(b/a)),\,\mu >0,\,a>0\,}$

In the present case, ${\displaystyle \mu =1,\,a=\tau _{0}^{-1},\,b=\omega \,}$, so the problem continues:

${\displaystyle ={\frac {1}{\pi }}{\frac {\Gamma (1)}{\left(\tau _{0}^{-2}+\omega ^{2}\right)^{1/2}}}\cos(\arctan(\omega \tau _{0}))\,}$

It is true that ${\displaystyle \Gamma (1)=1\,}$ and ${\displaystyle \cos(\arctan(x))={\frac {1}{\sqrt {x^{2}+1}}}\,}$, so the problem continues:

${\displaystyle ={\frac {1}{\pi }}\left[(\tau _{0}^{-2}+\omega ^{2})(\omega ^{2}\tau _{0}^{2}+1)\right]^{-1/2}\,}$

${\displaystyle ={\frac {1}{\pi }}(\omega ^{2}+\tau _{0}^{-2}+\omega ^{4}\tau _{0}^{2}+\omega ^{2})^{-1/2}\,}$

${\displaystyle ={\frac {1}{\pi }}\left(\left[1+2\omega ^{2}\tau _{0}^{2}+\omega ^{4}\tau _{0}^{4}\right]{\frac {1}{\tau _{0}^{2}}}\right)^{-1/2}\,}$

${\displaystyle ={\frac {1}{\pi }}\left(\left[1+\omega ^{2}\tau _{0}^{2}\right]^{2}{\frac {1}{\tau _{0}^{2}}}\right)^{-1/2}\,}$

${\displaystyle ={\frac {1}{\pi }}{\frac {\tau _{0}}{1+\omega ^{2}\tau _{0}^{2}}}\,}$