Question Number 85009 by mathmax by abdo last updated on 18/Mar/20
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{sh}\left({x}\right)}{dx}\:{with}\:{n}\:{integr}\:{natural} \\ $$
Commented by mathmax by abdo last updated on 18/Mar/20
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{sh}\left({x}\right)}{dx}\:\Rightarrow{A}_{{n}} =\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{e}^{{x}} −{e}^{−{x}} }{dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}} \:{x}^{{n}} }{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{{n}} \:{e}^{−{x}} \left(\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{kx}} \right){dx} \\ $$$$=\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−\left(\mathrm{2}{k}+\mathrm{1}\right){x}} \:{dx}\:=_{\left(\mathrm{2}{k}+\mathrm{1}\right){x}={t}} \:\:\mathrm{2}\sum_{{k}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\left(\frac{{t}}{\mathrm{2}{k}+\mathrm{1}}\right)^{{n}} \:{e}^{−{t}} \:\frac{{dt}}{\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} \:\:{t}^{{n}} \:{e}^{−{t}} \:{dt}\:=\mathrm{2}\Gamma\left({n}+\mathrm{1}\right)\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$${we}\:{have}\:\xi\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{x}} }\:\Rightarrow\xi\left({n}+\mathrm{1}\right)=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} } \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)^{{n}+\mathrm{1}} }+\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} }\:+\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$$\Rightarrow\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\right)\xi\left({n}+\mathrm{1}\right)\:\Rightarrow \\ $$$${A}_{{n}} =\mathrm{2}\Gamma\left({n}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\right)\xi\left({n}+\mathrm{1}\right) \\ $$$$=\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\left({n}!\right)\xi\left({n}+\mathrm{1}\right) \\ $$$$ \\ $$
Answered by mind is power last updated on 18/Mar/20
$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{2}{x}^{{n}} {e}^{{x}} {dx}}{\left({e}^{{x}} −\mathrm{1}\right)\left({e}^{{x}} +\mathrm{1}\right)} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{n}} {dx}}{{e}^{{x}} −\mathrm{1}}+\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{n}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{n}} {e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} {x}^{{n}} \underset{{k}\geqslant\mathrm{0}} {\sum}{e}^{−\left({k}+\mathrm{1}\right){x}} {dx}+\int_{\mathrm{0}} ^{+\infty} {x}^{{n}} .\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {e}^{−\left(\mathrm{1}+{k}\right){x}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{+\infty} {x}^{{n}} {e}^{−\left({k}+\mathrm{1}\right){x}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{+\infty} \frac{{y}^{{n}} }{\left({k}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{e}^{−{y}} {dy} \\ $$$$=\Gamma\left({n}+\mathrm{1}\right)\zeta\left({n}+\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{+\infty} {x}^{{n}} .\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {e}^{−\left(\mathrm{1}+{k}\right){x}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{+\infty} \frac{{y}^{{n}} }{\left(\mathrm{1}+{k}\right)^{{n}+\mathrm{1}} }\left(−\mathrm{1}\right)^{{k}} {e}^{−{y}} {dy} \\ $$$$=\Gamma\left({n}+\mathrm{1}\right)\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} .\frac{\mathrm{1}}{\left(\mathrm{1}+{k}\right)^{{n}+\mathrm{1}} } \\ $$$$\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{1}+{k}\right)^{{n}+\mathrm{1}} }=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}{k}\right)^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(\mathrm{2}+\mathrm{2}{k}\right)^{{n}+\mathrm{1}} }=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\zeta\left({n}+\mathrm{1}\right) \\ $$$$=\Gamma\left({n}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\zeta\left({n}+\mathrm{1}\right) \\ $$$$=\Gamma\left({n}+\mathrm{1}\right)\left\{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right\}\zeta\left({n}+\mathrm{1}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{n}} }{{sh}\left({x}\right)}{dx} \\ $$
Commented by abdomathmax last updated on 18/Mar/20
$${thank}\:{you}\:{sir}. \\ $$
Commented by mind is power last updated on 18/Mar/20
$${withe}\:{pleasur} \\ $$