Question Number 114681 by Dwaipayan Shikari last updated on 20/Sep/20
$${Prove}\:{that}\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}{dx}={n}!\zeta\left({n}+\mathrm{1}\right) \\ $$
Answered by mnjuly1970 last updated on 20/Sep/20
$$\:\:\:\:\:\:\:\:\:\:{proof}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Omega=\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}{dx}\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{n}} {e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }{dx}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\infty} \underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}} {e}^{−{mx}} {dx}=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} {x}^{{n}} {e}^{−{mx}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\overset{\langle{mx}\:=\:{t}\:\rangle} {=}\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\left({n}+\mathrm{1}\right)−\mathrm{1}} \:{e}^{−{t}} }{{m}^{{n}+\mathrm{1}} }{dt}\:=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)}{{m}^{{n}+\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\Gamma\left({n}+\mathrm{1}\right).\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{m}^{{n}+\mathrm{1}} }=\:{n}!\:\zeta\left({n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\checkmark\:{note}\:\left(\mathrm{1}\right)\:::\:\:\Gamma\left({n}+\mathrm{1}\right)={n}\Gamma\left({n}\right)={n}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\checkmark\:\:{note}\:\left(\mathrm{2}\right)\:::\:\:\zeta\left({s}\right)\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{m}.{n}.\:{july}\:\mathrm{1970}#…\:\: \\ $$$$\: \\ $$
Commented by Dwaipayan Shikari last updated on 20/Sep/20
$${Great}\:{sir}! \\ $$
Commented by mnjuly1970 last updated on 20/Sep/20
$${p}.{b}.{h}.{y}\:{you}\:{are}\:{welcom} \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by mathdave last updated on 20/Sep/20
$${solution}\: \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} {x}^{{n}} }{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$${series}\:{of} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{x}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{kx}} =\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−{x}\left({k}−\mathrm{1}\right)} =\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−{kx}+{x}} \\ $$$${I}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{n}} {e}^{−{kx}+{x}} {dx}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{−{kx}} \\ $$$${let}\:{y}={kx},{dx}=\frac{{dy}}{{k}} \\ $$$${I}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} \left(\frac{{y}}{{k}}\right)^{{n}} \frac{{dy}}{{k}}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} {y}^{{n}} {dy} \\ $$$${I}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {y}^{\left({n}+\mathrm{1}\right)−\mathrm{1}} {e}^{−{y}} {dy} \\ $$$${I}=\zeta\left({n}+\mathrm{1}\right)\Gamma\left({n}+\mathrm{1}\right)\:\:\:\:\:{but}\:\Gamma\left({n}+\mathrm{1}\right)={n}! \\ $$$$\because\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}={n}!\zeta\left({n}+\mathrm{1}\right)\:\:\:\:\:\:{Q}.{E}.{D} \\ $$$${by}\:{mathdave} \\ $$
Answered by Aziztisffola last updated on 20/Sep/20
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} }\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{e}^{{x}} }\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−\mathrm{k}{x}} \:{dx} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\int_{\mathrm{0}} ^{\:\infty} {x}^{{n}} {e}^{−{kx}−{x}} {dx} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\int_{\mathrm{0}} ^{\:\infty} {x}^{{n}} {e}^{−\left({k}+\mathrm{1}\right){x}} {dx} \\ $$$$\mathrm{let}\:\mathrm{t}=\left(\mathrm{k}+\mathrm{1}\right){x}\:\Rightarrow\mathrm{dt}=\left(\mathrm{k}+\mathrm{1}\right)\mathrm{d}{x}\: \\ $$$$\Rightarrow\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{t}}{\mathrm{k}+\mathrm{1}}\right)^{\mathrm{n}} {e}^{−\mathrm{t}} \frac{\mathrm{dt}}{\mathrm{k}+\mathrm{1}} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{t}^{\mathrm{n}} }{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }\:{e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\:\infty} \mathrm{t}^{\mathrm{n}} {e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} \mathrm{t}^{\mathrm{n}} {e}^{−\mathrm{t}} \mathrm{dt}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} \mathrm{t}^{\left(\mathrm{n}+\mathrm{1}\right)−\mathrm{1}} {e}^{−\mathrm{t}} \mathrm{dt}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{k}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$=\Gamma\left(\mathrm{n}+\mathrm{1}\right)\zeta\left(\mathrm{n}+\mathrm{1}\right) \\ $$$$=\mathrm{n}!\zeta\left(\mathrm{n}+\mathrm{1}\right) \\ $$