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Using Fourier Cosine integral prove that \[e^{-x}\cos{x=\frac{2}{\pi{}}\\int_0^{\infty{}}\frac{\left({\omega{}}^2+2\right)}{\left({\omega{}}^4+4\right)}.\\cos{\omega{}\ x\ d\omega{}}}\]
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$ Fourier \ Cosine\ Integral:\\[3ex] \displaystyle f(x)=\int_0^{\infty{}}A\left(w\right)\cos {wx\ dw} \\[2ex] \displaystyle where,\ A\left(w\right)=\frac{2}{\pi{}}\int_0^{\infty{}}f(x) \cos {wx\ }dx \\[2ex] \displaystyle A\left(w\right)=\frac{2}{\pi{}}\int_0^{\infty{}}e^{-x}\cos {x\ }\cos{wx\ }dx \\[2ex] \displaystyle \cos x\cos{wx=}\frac{1}{2}(\cosā”({x+wx)}+\cos{\left(x-wx\right)}) \\[2ex] \displaystyle A\left(w\right)=\frac{2}{2\pi{}}\int_0^{\infty{}}e^{-x}\cos{(x+wx)\ }dx+\ \frac{2}{2\pi{}}\int_0^{\infty{}}e^{-x}\cos{(x-wx)\ }dx \\[3ex] \displaystyle Using,\\[3ex] \displaystyle \int e^{ax}\cos{\left(bx\right)}dx=\frac{e^{ax}(a\cos{\left(bx\right)+b\sin{(bx)}})}{a^2+b^2}\\[2ex] \displaystyle A\left(w\right)=\frac{1}{\pi{}}\ {\left[\frac{e^{-x}\ (- \ cos{(1+w)x}+\left(1+w\right)\sin{(1+w)x)}}{{(-1)}^2+{\left(1+w\right)}^2}\right]}_0^{\infty{}}\\[2ex] \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle +\ \frac{1}{\pi{}}\ {\left[\frac{e^{-x}\ (-\ cos{(1-w)x}+\left(1-w\right)\sin{(1-w)x)}}{{(-1)}^2+{\left(1-w\right)}^2}\right]}_0^{\infty{}} \\[4ex] \displaystyle Now \ since \ e^{-\infty{}}=0 \\[2ex] \displaystyle A\left(w\right)=-\frac{1}{\pi{}}\ \left[\frac{e^0\left(-1\right)}{1+w^2+2w+1\ }+\frac{e^0\left(-1\right)}{1+w^2-2w+1}\right]\\[2ex] \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\pi{}}\left[\frac{1}{1+w^2+2w+1\ }+\frac{1}{1+w^2-2w+1}\right] \\[2ex] \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{1}{\pi{}}[\frac{2+w^2+2w+2+w^2-2w}{(2+w^2+2w)(2+w^2-2w)}] \\[2ex] \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2}{\pi{}}\left[\frac{w^2+2}{4+2w^2-4w+2w^2+w^4-2w^3+4w+2w^3-4w^2}\right] \\[2ex] \displaystyle \ \ A\left(w\right)\ \ \ =\frac{2}{\pi{}}\left[\frac{w^2+2\ }{w^4+4\ }\right] \\[2ex] \displaystyle f(x)=\int_0^{\infty{}}A\left(w\right)\cos{wx\ dw} \\[2ex] \displaystyle e^{-x}\cos{x\ }=\ \int_0^{\infty{}}\frac{2}{\pi{}}\left[\frac{w^2+2\ }{w^4+4\ }\right]\ \cos{wx\ dw}\ \ \ \ \\[2ex] \displaystyle e^{-x}\cos{x\ }=\frac{2}{\pi{}}\ \int_0^{\infty{}}\left[\frac{w^2+2\ }{w^4+4\ }\right]\ \cos{wx\ dw}\ \ \ \ \\[2ex]$

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