Question Number 198695 by universe last updated on 23/Oct/23
$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{at}} −\mathrm{e}^{−\mathrm{at}} }{\mathrm{e}^{\pi\mathrm{t}} −\mathrm{e}^{−\pi\mathrm{t}} }\:\mathrm{dt}\:\:\:=\:\:\:?? \\ $$
Answered by witcher3 last updated on 25/Oct/23
![existe ⇔a∈]−π,π[ f(a)=∫_0 ^∞ ((e^(at) −e^(−at) )/(e^(πt) −e^(−πt) ))dt,f(−a)=−f(a) ∀t∈[0,a[ ∫_0 ^∞ ((e^(at) −e^(−at) )/(e^(πt) −e^(−πt) ))dt=∫_0 ^∞ (e^(at) −e^(−at) )e^(−πt) Σe^(−2nπt) dt =Σ_(n≥0) ∫_0 ^∞ e^(−(π−a+2nπ)t) −e^(−(a+π+2nπ)t) dt =Σ_(n≥0) ((1/(π−a+2nπ))−(1/(a+π+2nπ))) =(1/(2π))Σ_(n≥0) ((1/(n+(1/2)−(a/(2π))))−(1/(n+(1/2)+(a/(2π))))) =(1/(2π))Σ_(n≥0) ((1/(n+1))−(1/(n+(1/2)+(a/(2π))))−((1/(n+1))−(1/(n+(1/2)−(a/(2π)))))) =(1/(2π))Σ_(n≥0) (1/(n+1))−(1/(n+(1/2)+(a/(2π))))−(1/(2π))Σ_(n≥0) (1/(n+1))−(1/(n+(1/2)−(a/(2π)))) =(1/(2π))(Ψ((1/2)+(a/(2π)))−Ψ((1/2)−(a/(2π)))) =(1/(2π))(Ψ(1−((1/2)−(a/(2π))))−Ψ((1/2)−(a/(2π)))) =(1/(2π))(.πcot(π((1/2)−(a/(2π))))=((tan((a/2)))/2) ∫_0 ^∞ ((e^(ax) −e^(−ax) )/(e^(πx) −e^(−πx) ))dx=((tan ((a/2)))/2)](https://www.tinkutara.com/question/Q198899.png)
$$\left.\mathrm{existe}\:\Leftrightarrow\mathrm{a}\in\right]−\pi,\pi\left[\right. \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{at}} −\mathrm{e}^{−\mathrm{at}} }{\mathrm{e}^{\pi\mathrm{t}} −\mathrm{e}^{−\pi\mathrm{t}} }\mathrm{dt},\mathrm{f}\left(−\mathrm{a}\right)=−\mathrm{f}\left(\mathrm{a}\right) \\ $$$$\forall\mathrm{t}\in\left[\mathrm{0},\mathrm{a}\left[\right.\right. \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{at}} −\mathrm{e}^{−\mathrm{at}} }{\mathrm{e}^{\pi\mathrm{t}} −\mathrm{e}^{−\pi\mathrm{t}} }\mathrm{dt}=\int_{\mathrm{0}} ^{\infty} \left(\mathrm{e}^{\mathrm{at}} −\mathrm{e}^{−\mathrm{at}} \right)\mathrm{e}^{−\pi\mathrm{t}} \Sigma\mathrm{e}^{−\mathrm{2n}\pi\mathrm{t}} \mathrm{dt} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\left(\pi−\mathrm{a}+\mathrm{2n}\pi\right)\mathrm{t}} −\mathrm{e}^{−\left(\mathrm{a}+\pi+\mathrm{2n}\pi\right)\mathrm{t}} \mathrm{dt} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\pi−\mathrm{a}+\mathrm{2n}\pi}−\frac{\mathrm{1}}{\mathrm{a}+\pi+\mathrm{2n}\pi}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{2}\pi}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{a}}{\mathrm{2}\pi}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{a}}{\mathrm{2}\pi}}−\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{2}\pi}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{a}}{\mathrm{2}\pi}}−\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{2}\pi}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{a}}{\mathrm{2}\pi}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{2}\pi}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\left(\Psi\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{2}\pi}\right)\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{2}\pi}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\left(.\pi\mathrm{cot}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{2}\pi}\right)\right)=\frac{\mathrm{tan}\left(\frac{\mathrm{a}}{\mathrm{2}}\right)}{\mathrm{2}}\right. \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{ax}} −\mathrm{e}^{−\mathrm{ax}} }{\mathrm{e}^{\pi\mathrm{x}} −\mathrm{e}^{−\pi\mathrm{x}} }\mathrm{dx}=\frac{\mathrm{tan}\:\left(\frac{\mathrm{a}}{\mathrm{2}}\right)}{\mathrm{2}} \\ $$