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Question Number 78269 by msup trace by abdo last updated on 15/Jan/20

$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{cos\left(ax\right)}{x^2+a^2}dxwith$$$$a>0find{\int }_1^{2}f\left(a\right)da$$

Commented by mathmax by abdo last updated on 15/Jan/20

$$wehave2f\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(ax\right)}{x^2+a^2}dx=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{iax}}{x^2+a^2}dx\right)$$$$letW\left(z\right)=\frac{e^{iaz}}{z^2+a^2}\Rightarrow W\left(z\right)=\frac{e^{iaz}}{(z-ia)(z+ia)}sothepolesofWare$$$$iaand-ia\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi Res(W,ia)$$$$=2i\pi \times \frac{e^{ia\left(ia\right)}}{2ia}=\frac{\pi }{a}{e}^{-a^2}=2f\left(a\right)\Rightarrow f\left(a\right)=\frac{\pi }{2a}{e}^{-a^2}\Rightarrow$$$${\int }_1^{2}f\left(a\right)da=\frac{\pi }{2}{\int }_1^2\frac{e^{-a^2}}{a}daandaformofserie$$$$e^{-a^2}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-a^2)}^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{a}^{2n}}{n!}\Rightarrow$$$$\frac{e^{-a^2}}{a}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}{a}^{2n-1}\Rightarrow {\int }_1^2\frac{e^{-a^2}}{a}da$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}{\left[\frac{1}{2n}{a}^{2n}\right]}_1^2=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)n!}(2^{2n}-1)\Rightarrow$$$${\int }_1^{2}f\left(a\right)da=\frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)n!}(4^n-1)$$

Answered by mind is power last updated on 15/Jan/20

$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{+\mathrm{\infty }}\frac{\cos \left(\mathrm{ax}\right)}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}$$$$=\mathrm{Re}\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{iax}}}{{\mathrm{x}}^2+{\mathrm{a}}^2}$$$$=\mathrm{Re}\left\{\mathrm{i}\pi \mathrm{Res}(\frac{{\mathrm{e}}^{\mathrm{iax}}}{{\mathrm{x}}^2+{\mathrm{a}}^2},\mathrm{x}=\mathrm{ia})\right\}$$$$=\mathrm{i}\pi .\frac{{\mathrm{e}}^{-{\mathrm{a}}^2}}{2\mathrm{ia}}=\frac{\pi }{{2\mathrm{ae}}^{-\mathrm{a}}}$$$${\int }_1^2\frac{\pi {\mathrm{e}}^{\mathrm{a}}}{2\mathrm{a}}=\frac{\pi }{2}\int \frac{{\mathrm{e}}^{\mathrm{a}}}{\mathrm{a}}\mathrm{da}$$$$\mathrm{use}\mathrm{E}\left(\mathrm{x}\right)$$$$$$$$$$

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