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Question Number 198695 by universe last updated on 23/Oct/23

$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{at}}-{\mathrm{e}}^{-\mathrm{at}}}{{\mathrm{e}}^{\pi \mathrm{t}}-{\mathrm{e}}^{-\pi \mathrm{t}}}\mathrm{dt}=??$$

Answered by witcher3 last updated on 25/Oct/23

$$\mathrm{existe}\iff \mathrm{a}\in ]-\pi ,\pi [$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{at}}-{\mathrm{e}}^{-\mathrm{at}}}{{\mathrm{e}}^{\pi \mathrm{t}}-{\mathrm{e}}^{-\pi \mathrm{t}}}\mathrm{dt},\mathrm{f}(-\mathrm{a})=-\mathrm{f}\left(\mathrm{a}\right)$$$$\mathrm{\forall }\mathrm{t}\in [0,\mathrm{a}[$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{at}}-{\mathrm{e}}^{-\mathrm{at}}}{{\mathrm{e}}^{\pi \mathrm{t}}-{\mathrm{e}}^{-\pi \mathrm{t}}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}({\mathrm{e}}^{\mathrm{at}}-{\mathrm{e}}^{-\mathrm{at}}){\mathrm{e}}^{-\pi \mathrm{t}}\mathrm{\Sigma }{\mathrm{e}}^{-2\mathrm{n}\pi \mathrm{t}}\mathrm{dt}$$$$=\sum _{\mathrm{n}⩾0}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\pi -\mathrm{a}+2\mathrm{n}\pi )\mathrm{t}}-{\mathrm{e}}^{-(\mathrm{a}+\pi +2\mathrm{n}\pi )\mathrm{t}}\mathrm{dt}$$$$=\sum _{\mathrm{n}⩾0}(\frac{1}{\pi -\mathrm{a}+2\mathrm{n}\pi }-\frac{1}{\mathrm{a}+\pi +2\mathrm{n}\pi })$$$$=\frac{1}{2\pi }\sum _{\mathrm{n}⩾0}(\frac{1}{\mathrm{n}+\frac{1}{2}-\frac{\mathrm{a}}{2\pi }}-\frac{1}{\mathrm{n}+\frac{1}{2}+\frac{\mathrm{a}}{2\pi }})$$$$=\frac{1}{2\pi }\sum _{\mathrm{n}⩾0}(\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+\frac{1}{2}+\frac{\mathrm{a}}{2\pi }}-(\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+\frac{1}{2}-\frac{\mathrm{a}}{2\pi }}))$$$$=\frac{1}{2\pi }\sum _{\mathrm{n}⩾0}\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+\frac{1}{2}+\frac{\mathrm{a}}{2\pi }}-\frac{1}{2\pi }\sum _{\mathrm{n}⩾0}\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+\frac{1}{2}-\frac{\mathrm{a}}{2\pi }}$$$$=\frac{1}{2\pi }(\mathrm{\Psi }(\frac{1}{2}+\frac{\mathrm{a}}{2\pi })-\mathrm{\Psi }(\frac{1}{2}-\frac{\mathrm{a}}{2\pi }))$$$$=\frac{1}{2\pi }(\mathrm{\Psi }(1-(\frac{1}{2}-\frac{\mathrm{a}}{2\pi }))-\mathrm{\Psi }(\frac{1}{2}-\frac{\mathrm{a}}{2\pi }))$$$$=\frac{1}{2\pi }(.\pi \cot \left(\pi (\frac{1}{2}-\frac{\mathrm{a}}{2\pi })\right)=\frac{\tan \left(\frac{\mathrm{a}}{2}\right)}{2}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ax}}-{\mathrm{e}}^{-\mathrm{ax}}}{{\mathrm{e}}^{\pi \mathrm{x}}-{\mathrm{e}}^{-\pi \mathrm{x}}}\mathrm{dx}=\frac{\tan \left(\frac{\mathrm{a}}{2}\right)}{2}$$

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