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Question Number 139092 by mathdanisur last updated on 22/Apr/21

$$\mathrm{\Omega }=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{xsinh\left(\pi x\right)e^{-x^2}}{1+x^2}dx$$

Answered by mathmax by abdo last updated on 23/Apr/21

$$\mathrm{i}\mathrm{give}\mathrm{this}\mathrm{soution}\mathrm{but}\mathrm{not}\mathrm{sure}!$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{xsh}\left(\pi \mathrm{x}\right){\mathrm{e}}^{-{\mathrm{x}}^2}}{{\mathrm{x}}^2+1}\mathrm{dx}\Rightarrow \mathrm{\Phi }=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{xsh}\left(\pi \mathrm{x}\right)}{{\mathrm{x}}^2+1}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{zsh}\left(\pi \mathrm{z}\right){\mathrm{e}}^{-{\mathrm{z}}^2}}{{\mathrm{z}}^2+1}\Rightarrow \mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{zsh}\left(\pi \mathrm{z}\right){\mathrm{e}}^{-{\mathrm{z}}^2}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\mathrm{w},\mathrm{i})[=2\mathrm{i}\pi \times \frac{\mathrm{ish}\left(\pi \mathrm{i}\right){\mathrm{e}}^{-(-1)}}{2\mathrm{i}}=\mathrm{i}\pi \mathrm{sh}\left(\pi \mathrm{i}\right)\mathrm{e}$$$$\mathrm{but}\mathrm{sh}\left(\pi \mathrm{i}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\left(\pi \mathrm{i}\right)}-{\mathrm{e}}^{-\mathrm{i}\left(\pi \mathrm{i}\right)}}{2\mathrm{i}}=\frac{{\mathrm{e}}^{-\pi }-{\mathrm{e}}^{\pi }}{2\mathrm{i}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{i}\pi \left(\frac{{\mathrm{e}}^{-\pi }-{\mathrm{e}}^{\pi }}{2\mathrm{i}}\right)\mathrm{e}=\pi ({\mathrm{e}}^{1-\pi }-{\mathrm{e}}^{1+\pi })\Rightarrow$$$$\mathrm{\Phi }=\frac{\pi }{2}({\mathrm{e}}^{1-\pi }-{\mathrm{e}}^{1+\pi })$$

Commented by mathdanisur last updated on 23/Apr/21

$$thaksthanksSir$$

Commented by mathdanisur last updated on 23/Apr/21

$$Sir,sinh\left(i\pi \right)=0?$$

Commented by mathmax by abdo last updated on 24/Apr/21

$$\mathrm{no}\mathrm{sir}\mathrm{its}\mathrm{not}\mathrm{like}\mathrm{sinx}...!$$

Commented by mathdanisur last updated on 24/Apr/21

$$Sir,omegamustbe>0.$$$$Youranswerisanegativenumber$$

Commented by mathdanisur last updated on 24/Apr/21

$$Sir,{it}^{\prime }snotlikesin,sin\left(ix\right)/i=sinh?$$

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