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Question Number 85346 by naka3546 last updated on 21/Mar/20

$$\underset{0}{\int }\stackrel{\mathrm{\infty }}{}\frac{\cos \left(\pi x\right)}{{\pi }^2+x^2}dx=?$$

Commented by abdomathmax last updated on 21/Mar/20

$$A={\int }_0^{\mathrm{\infty }}\frac{cos\left(\pi x\right)}{x^2+{\pi }^2}dx\Rightarrow 2A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(\pi x\right)}{x^2+{\pi }^2}dx$$$$=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\pi x}}{x^2+{\pi }^2}dx\right)let\phi \left(z\right)=\frac{e^{i\pi z}}{z^2+{\pi }^2}$$$$wehave\phi \left(z\right)=\frac{e^{i\pi z}}{(z-i\pi )(z+i\pi )}$$$$residustheiremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i\pi )$$$$=2i\pi \frac{e^{i\pi \left(i\pi \right)}}{2i\pi }=e^{-{\pi }^2}\Rightarrow A=\frac{e^{-{\pi }^2}}{2}$$

Answered by M±th+et£s last updated on 21/Mar/20

$$f\left(p\right)={\int }_0^{\mathrm{\infty }}\frac{cos\left(p\pi x\right)}{{\pi }^2+x^2}dx\Rightarrow \Rightarrow \Rightarrow \lfloor \left(f\left(p\right)\right)={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{cos\left(p\pi x\right)}{{\pi }^2+x^2}dxdp$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{{\pi }^2+x^2}{\int }_0^{\mathrm{\infty }}{e}^{-ps}cos\left(p\pi x\right)dxdp$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{({\pi }^2+x^2)}\left(\frac{s}{s^2+{\pi }^2{x}^2}\right)dx$$$$=\frac{s}{s-{\pi }^4}{\int }_0^{\mathrm{\infty }}\frac{s^2+{\pi }^2{x}^2-{\pi }^4-{\pi }^2{x}^2}{({\pi }^2+x^2)(s^2+{\pi }^2{x}^2)}dx$$$$\frac{s}{s^2-{\pi }^4}\int (\frac{1}{{\pi }^2+x^2}-\frac{{\pi }^2}{s^2+{\pi }^2{x}^2})dx$$$$\frac{s}{s^2-{\pi }^4}{[\frac{1}{\pi }{tan}^{-1}\frac{x}{\pi }-\frac{\pi }{s}{tan}^{-1}\frac{\pi x}{s}]}_0^{\mathrm{\infty }}$$$$\frac{s}{s^2-{\pi }^4}[\frac{1}{2}-\frac{{\pi }^2}{2s}]$$$$\frac{s}{s^2-{\pi }^4}\left(\frac{s-{\pi }^2}{2s}\right)=\frac{1}{2(s+{\pi }^2)}$$$$f\left(p\right)={\lfloor }^{-1}\left(\frac{1}{s+{\pi }^2}\right)$$$$=\frac{1}{2}{e}^{-{\pi }^2}$$$$$$$$$$

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