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Question Number 152240 by mnjuly1970 last updated on 26/Aug/21
prove that.. csch (x)= (1/x) + Σ_(n=1) ^∞ ((2.(−1)^( n) x)/(n^( 2) π^( 2) + x^( 2) )) then find: Ω := ∫_0 ^( ∞) ((cosh (x )−(1/x))/x) dx=−ln(2)....■
$$ \\ $$$$\:\:{prove}\:{that}.. \\ $$$$\:\:\:{csch}\:\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}.\left(−\mathrm{1}\right)^{\:{n}} \:{x}}{{n}^{\:\mathrm{2}} \pi^{\:\mathrm{2}} +\:{x}^{\:\mathrm{2}} } \\ $$$$\:\:\:{then}\:{find}: \\ $$$$\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cosh}\:\left({x}\:\right)−\frac{\mathrm{1}}{{x}}}{{x}}\:{dx}=−{ln}\left(\mathrm{2}\right)….\blacksquare \\ $$
Answered by Kamel last updated on 27/Aug/21
Ω=∫_0 ^(+∞) ((cosech(x)−(1/x))/x)dx=∫_0 ^(+∞) ∫_0 ^(+∞) (((2e^(−x) )/(1−e^(−2x) ))−(1/x))e^(−tx) dxdt =^(IBP) ∫_0 ^(+∞) (Ln(2)+t∫_0 ^(+∞) (Ln(((1−e^(−x) )/x))−Ln(1+e^(−x) ))e^(−xt) dx)dt =∫_0 ^(+∞) (Ln(2)+t(2∫_0 ^1 Ln(1−u)u^(t−1) du−(1/2)∫_0 ^1 Ln(1−u)u^((t/2)−1) du+((γ+Ln(t))/t)))dt =∫_0 ^(+∞) (Ln(2)+t(−(2/t)∫_0 ^1 ((u^t −1)/(u−1))du+(1/t)∫_0 ^1 ((u^(t/2) −1)/(u−1))du+((γ+Ln(t))/t)))dt =∫_0 ^(+∞) (Ln(t)+Ψ((t/2)+1)−2Ψ(t+1)+Ln(2))dt =lim_(t→+∞) (tLn(t)−t+2Ln(Γ((t/2)+1))−2LnΓ(t+1)+tLn(2)) =lim_(t→+∞) Ln(((t^t e^(−t) Γ^2 ((t/2)+1)2^t )/(Γ^2 (t+1))))=lim_(t→+∞) Ln(((t^(2t) πte^(−2t) )/(t^(2t) 2πte^(−2t) )))=−Ln(2) ∴ ∫_0 ^(+∞) ((cosech(x)−(1/x))/x)dx=−Ln(2) KAMEL BENAICHA
$$ \\ $$$$\Omega=\int_{\mathrm{0}} ^{+\infty} \frac{{cosech}\left({x}\right)−\frac{\mathrm{1}}{{x}}}{{x}}{dx}=\int_{\mathrm{0}} ^{+\infty} \int_{\mathrm{0}} ^{+\infty} \left(\frac{\mathrm{2}{e}^{−{x}} }{\mathrm{1}−{e}^{−\mathrm{2}{x}} }−\frac{\mathrm{1}}{{x}}\right){e}^{−{tx}} {dxdt} \\ $$$$\:\:\overset{{IBP}} {=}\int_{\mathrm{0}} ^{+\infty} \left({Ln}\left(\mathrm{2}\right)+{t}\int_{\mathrm{0}} ^{+\infty} \left({Ln}\left(\frac{\mathrm{1}−{e}^{−{x}} }{{x}}\right)−{Ln}\left(\mathrm{1}+{e}^{−{x}} \right)\right){e}^{−{xt}} {dx}\right){dt} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{+\infty} \left({Ln}\left(\mathrm{2}\right)+{t}\left(\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}\left(\mathrm{1}−{u}\right){u}^{{t}−\mathrm{1}} {du}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}\left(\mathrm{1}−{u}\right){u}^{\frac{{t}}{\mathrm{2}}−\mathrm{1}} {du}+\frac{\gamma+{Ln}\left({t}\right)}{{t}}\right)\right){dt} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{+\infty} \left({Ln}\left(\mathrm{2}\right)+{t}\left(−\frac{\mathrm{2}}{{t}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{{t}} −\mathrm{1}}{{u}−\mathrm{1}}{du}+\frac{\mathrm{1}}{{t}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\frac{{t}}{\mathrm{2}}} −\mathrm{1}}{{u}−\mathrm{1}}{du}+\frac{\gamma+{Ln}\left({t}\right)}{{t}}\right)\right){dt} \\ $$$$\:\:=\int_{\mathrm{0}} ^{+\infty} \left({Ln}\left({t}\right)+\Psi\left(\frac{{t}}{\mathrm{2}}+\mathrm{1}\right)−\mathrm{2}\Psi\left({t}+\mathrm{1}\right)+{Ln}\left(\mathrm{2}\right)\right){dt} \\ $$$$\:=\underset{{t}\rightarrow+\infty} {{lim}}\left({tLn}\left({t}\right)−{t}+\mathrm{2}{Ln}\left(\Gamma\left(\frac{{t}}{\mathrm{2}}+\mathrm{1}\right)\right)−\mathrm{2}{Ln}\Gamma\left({t}+\mathrm{1}\right)+{tLn}\left(\mathrm{2}\right)\right) \\ $$$$\:=\underset{{t}\rightarrow+\infty} {{lim}Ln}\left(\frac{{t}^{{t}} {e}^{−{t}} \Gamma^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}+\mathrm{1}\right)\mathrm{2}^{{t}} }{\Gamma^{\mathrm{2}} \left({t}+\mathrm{1}\right)}\right)=\underset{{t}\rightarrow+\infty} {{lim}Ln}\left(\frac{{t}^{\mathrm{2}{t}} \pi{te}^{−\mathrm{2}{t}} }{{t}^{\mathrm{2}{t}} \mathrm{2}\pi{te}^{−\mathrm{2}{t}} }\right)=−{Ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \frac{\boldsymbol{{cosech}}\left(\boldsymbol{{x}}\right)−\frac{\mathrm{1}}{\boldsymbol{{x}}}}{\boldsymbol{{x}}}\boldsymbol{{dx}}=−\boldsymbol{{Ln}}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{KAMEL}}\:\boldsymbol{{BENAICHA}} \\ $$

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