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Question Number 156915 by mnjuly1970 last updated on 17/Oct/21

Answered by qaz last updated on 07/Nov/21

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{({\pi }^2+{\mathrm{x}}^2)\cosh \mathrm{x}}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{x}}}{({\mathrm{e}}^{2\mathrm{x}}+1)({\pi }^2+{\mathrm{x}}^2)}\mathrm{dx}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{2\mathrm{x}}+1}\cdot \frac{1}{\pi }\mathrm{\Re }{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\pi -\mathrm{ix})\mathrm{y}}\mathrm{dydx}$$$$=\frac{1}{\pi }\mathrm{\Re }{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\pi \mathrm{y}}\mathrm{dy}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{x}(1+\mathrm{iy})}}{{\mathrm{e}}^{2\mathrm{x}}+1}\mathrm{dx}$$$$=\frac{1}{2\pi }\mathrm{\Re }{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\pi \mathrm{y}}\mathrm{dy}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{\mathrm{iy}-1}{2}}}{\mathrm{t}+1}\mathrm{dt}.......{\mathrm{e}}^{\mathrm{x}}\to \mathrm{t}$$$$=\frac{1}{2}\mathrm{\Re }{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\pi \mathrm{y}}}{\sin \left(\frac{\mathrm{iy}+1}{2}\pi \right)}\mathrm{dy}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\pi \mathrm{y}}}{\cosh \frac{\pi \mathrm{y}}{2}}\mathrm{dy}$$$$=\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{{2\mathrm{e}}^{-3\mathrm{y}}}{1+{\mathrm{e}}^{-2\mathrm{y}}}\mathrm{dy}$$$$=\frac{2}{\pi }\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(2\mathrm{n}+3)\mathrm{y}}\mathrm{dy}$$$$=\frac{2}{\pi }\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+3}$$$$=\frac{2}{\pi }(1-\frac{\pi }{4})$$$$=\frac{2}{\pi }-\frac{1}{2}$$

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