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Question Number 114681 by Dwaipayan Shikari last updated on 20/Sep/20

Prove that ∫_0 ^∞ (x^n /(e^x −1))dx=n!ζ(n+1)

$${Prove}\:{that}\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}{dx}={n}!\zeta\left({n}+\mathrm{1}\right) \\ $$

Answered by mnjuly1970 last updated on 20/Sep/20

proof:: Ω= ∫_0 ^( ∞) (x^n /(e^x −1))dx =∫_0 ^( ∞) ((x^n e^(−x) )/(1−e^(−x) ))dx = ∫_0 ^( ∞) Σ_(m=1) ^∞ x^n e^(−mx) dx=Σ_(m=1) ^∞ ∫_0 ^( ∞) x^n e^(−mx) dx =^(⟨mx = t ⟩) Σ_(m=1) ^∞ ∫_0 ^( ∞) ((t^((n+1)−1) e^(−t) )/m^(n+1) )dt =Σ_(m=1) ^∞ ((Γ(n+1))/m^(n+1) ) Ω=Γ(n+1).Σ_(m=1) ^∞ (1/m^(n+1) )= n! ζ(n+1) ✓ note (1) :: Γ(n+1)=nΓ(n)=n! ✓ note (2) :: ζ(s) =Σ_(n=1) ^∞ (1/n^s ) ...m.n. july 1970#...

$$\:\:\:\:\:\:\:\:\:\:{proof}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Omega=\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}{dx}\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{n}} {e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }{dx}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\infty} \underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}} {e}^{−{mx}} {dx}=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} {x}^{{n}} {e}^{−{mx}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\overset{\langle{mx}\:=\:{t}\:\rangle} {=}\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\left({n}+\mathrm{1}\right)−\mathrm{1}} \:{e}^{−{t}} }{{m}^{{n}+\mathrm{1}} }{dt}\:=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)}{{m}^{{n}+\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\Gamma\left({n}+\mathrm{1}\right).\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{m}^{{n}+\mathrm{1}} }=\:{n}!\:\zeta\left({n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\checkmark\:{note}\:\left(\mathrm{1}\right)\:::\:\:\Gamma\left({n}+\mathrm{1}\right)={n}\Gamma\left({n}\right)={n}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\checkmark\:\:{note}\:\left(\mathrm{2}\right)\:::\:\:\zeta\left({s}\right)\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.\:{july}\:\mathrm{1970}#...\:\: \\ $$$$\: \\ $$

Commented by Dwaipayan Shikari last updated on 20/Sep/20

Great sir!

$${Great}\:{sir}! \\ $$

Commented by mnjuly1970 last updated on 20/Sep/20

p.b.h.y you are welcom

$${p}.{b}.{h}.{y}\:{you}\:{are}\:{welcom} \\ $$

Commented by Tawa11 last updated on 06/Sep/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Answered by mathdave last updated on 20/Sep/20

solution let I=∫_0 ^∞ (x^n /(e^x −1))dx=∫_0 ^∞ ((e^(−x) x^n )/(1−e^(−x) ))dx series of (1/(1−e^(−x) ))=Σ_(k=0) ^∞ e^(−kx) =Σ_(k=1) ^∞ e^(−x(k−1)) =Σ_(k=1) ^∞ e^(−kx+x) I=Σ_(k=1) ^∞ ∫_0 ^∞ e^(−x) x^n e^(−kx+x) dx=Σ_(k=1) ^∞ ∫_0 ^∞ x^n e^(−kx) let y=kx,dx=(dy/k) I=Σ_(k=1) ^∞ ∫_0 ^∞ e^(−y) ((y/k))^n (dy/k)=Σ_(k=1) ^∞ (1/k^(n+1) )∫_0 ^∞ e^(−y) y^n dy I=Σ_(k=1) ^∞ (1/k^(n+1) )∫_0 ^∞ y^((n+1)−1) e^(−y) dy I=ζ(n+1)Γ(n+1) but Γ(n+1)=n! ∵∫_0 ^∞ (x^n /(e^x −1))=n!ζ(n+1) Q.E.D by mathdave

$${solution}\: \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} {x}^{{n}} }{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$${series}\:{of} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{x}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{kx}} =\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−{x}\left({k}−\mathrm{1}\right)} =\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−{kx}+{x}} \\ $$$${I}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{n}} {e}^{−{kx}+{x}} {dx}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{−{kx}} \\ $$$${let}\:{y}={kx},{dx}=\frac{{dy}}{{k}} \\ $$$${I}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} \left(\frac{{y}}{{k}}\right)^{{n}} \frac{{dy}}{{k}}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} {y}^{{n}} {dy} \\ $$$${I}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {y}^{\left({n}+\mathrm{1}\right)−\mathrm{1}} {e}^{−{y}} {dy} \\ $$$${I}=\zeta\left({n}+\mathrm{1}\right)\Gamma\left({n}+\mathrm{1}\right)\:\:\:\:\:{but}\:\Gamma\left({n}+\mathrm{1}\right)={n}! \\ $$$$\because\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}={n}!\zeta\left({n}+\mathrm{1}\right)\:\:\:\:\:\:{Q}.{E}.{D} \\ $$$${by}\:{mathdave} \\ $$

Answered by Aziztisffola last updated on 20/Sep/20

∫_0 ^∞ (x^n /(e^x −1))dx=∫_0 ^∞ (x^n /e^x ) (1/(1−e^(−x) ))dx =∫_0 ^∞ (x^n /e^x ) Σ_(k=0) ^∞ e^(−kx) dx =Σ_(k=0) ^∞ ∫_0 ^( ∞) x^n e^(−kx−x) dx =Σ_(k=0) ^∞ ∫_0 ^( ∞) x^n e^(−(k+1)x) dx let t=(k+1)x ⇒dt=(k+1)dx ⇒ Σ_(k=0) ^∞ ∫_0 ^( ∞) ((t/(k+1)))^n e^(−t) (dt/(k+1)) =Σ_(k=0) ^∞ ∫_0 ^( ∞) (t^n /((k+1)^(n+1) )) e^(−t) dt = Σ_(k=0) ^∞ (1/((k+1)^(n+1) ))∫_0 ^( ∞) t^n e^(−t) dt =∫_0 ^( ∞) t^n e^(−t) dtΣ_(k=0) ^∞ (1/((k+1)^(n+1) )) =∫_0 ^( ∞) t^((n+1)−1) e^(−t) dtΣ_(k=1) ^∞ (1/((k)^(n+1) )) =Γ(n+1)ζ(n+1) =n!ζ(n+1)

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{e}^{{x}} }\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{e}^{{x}} }\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−\mathrm{k}{x}} \:{dx} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\int_{\mathrm{0}} ^{\:\infty} {x}^{{n}} {e}^{−{kx}−{x}} {dx} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\int_{\mathrm{0}} ^{\:\infty} {x}^{{n}} {e}^{−\left({k}+\mathrm{1}\right){x}} {dx} \\ $$$$\mathrm{let}\:\mathrm{t}=\left(\mathrm{k}+\mathrm{1}\right){x}\:\Rightarrow\mathrm{dt}=\left(\mathrm{k}+\mathrm{1}\right)\mathrm{d}{x}\: \\ $$$$\Rightarrow\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{t}}{\mathrm{k}+\mathrm{1}}\right)^{\mathrm{n}} {e}^{−\mathrm{t}} \frac{\mathrm{dt}}{\mathrm{k}+\mathrm{1}} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{t}^{\mathrm{n}} }{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }\:{e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\:\infty} \mathrm{t}^{\mathrm{n}} {e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} \mathrm{t}^{\mathrm{n}} {e}^{−\mathrm{t}} \mathrm{dt}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} \mathrm{t}^{\left(\mathrm{n}+\mathrm{1}\right)−\mathrm{1}} {e}^{−\mathrm{t}} \mathrm{dt}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{k}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$=\Gamma\left(\mathrm{n}+\mathrm{1}\right)\zeta\left(\mathrm{n}+\mathrm{1}\right) \\ $$$$=\mathrm{n}!\zeta\left(\mathrm{n}+\mathrm{1}\right) \\ $$

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