advanced-calculus-0-dx-x-5-e-1-x-1-

Question Number 131866 by mnjuly1970 last updated on 09/Feb/21

...advanced calculus... Ω=∫_0 ^( ∞) (dx/(x^5 (e^(1/x) −1)))=?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:\:{calculus}… \\ $$$$\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{dx}}{{x}^{\mathrm{5}} \left({e}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}\right)}=? \\ $$$$ \\ $$

Answered by mnjuly1970 last updated on 09/Feb/21

solution: Ω=^(x=(1/t)) ∫_0 ^( ∞) (t^5 /(e^t −1))∗(dt/t^2 ) =∫_0 ^( ∞) (t^3 /(e^t −1))dt=Γ(4)ζ(4)=(π^4 /(15)) note: ∫_0 ^( ∞) (x^(s−1) /(e^x −1))dx=Γ(s).ζ(s)

$$\:\:\:\:\:{solution}: \\ $$$$\:\:\:\Omega\overset{{x}=\frac{\mathrm{1}}{{t}}} {=}\:\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\mathrm{5}} }{{e}^{{t}} −\mathrm{1}}\ast\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\mathrm{3}} }{{e}^{{t}} −\mathrm{1}}{dt}=\Gamma\left(\mathrm{4}\right)\zeta\left(\mathrm{4}\right)=\frac{\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$$$\:\:\:{note}:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{s}−\mathrm{1}} }{{e}^{{x}} −\mathrm{1}}{dx}=\Gamma\left({s}\right).\zeta\left({s}\right) \\ $$

Answered by mathmax by abdo last updated on 10/Feb/21

Φ=∫_0 ^∞ (dx/(x^5 (e^(1/x) −1))) ⇒Φ=_((1/x)=t) −∫_0 ^∞ (t^5 /((e^t −1)))(−(dt/t^2 ))=∫_0 ^∞ (t^3 /(e^t −1))dt =∫_0 ^∞ ((e^(−t) t^3 )/(1−e^(−t) ))dt =∫_0 ^∞ t^3 e^(−t) Σ_(n=0) ^∞ e^(−nt) dt =Σ_(n=0) ^∞ ∫_0 ^∞ t^3 e^(−(n+1)t) dt =_((n+1)t=u) Σ_(n=0) ^∞ ∫_0 ^∞ (u^3 /((n+1)^3 ))e^(−u) (du/(n+1)) =Σ_(n=0) ^∞ (1/((n+1)^4 ))∫_0 ^∞ u^(4−1) e^(−u) du =ξ(4).Γ(4) =(π^4 /(90)).3! =(6/(90)).π^4 =((2.3)/(3.30))π^4 =(π^4 /(15))

$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{5}} \left(\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} −\mathrm{1}\right)}\:\Rightarrow\Phi=_{\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{t}} \:\:\:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{5}} }{\left(\mathrm{e}^{\mathrm{t}} −\mathrm{1}\right)}\left(−\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{t}} \:\mathrm{t}^{\mathrm{3}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{t}} }\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{t}} \:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{nt}} \:\mathrm{dt} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\left(\mathrm{n}+\mathrm{1}\right)\mathrm{t}} \:\mathrm{dt}\:=_{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{t}=\mathrm{u}} \:\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{\mathrm{3}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{e}^{−\mathrm{u}} \:\frac{\mathrm{du}}{\mathrm{n}+\mathrm{1}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{4}} }\int_{\mathrm{0}} ^{\infty} \:\mathrm{u}^{\mathrm{4}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{u}} \:\mathrm{du}\:\:=\xi\left(\mathrm{4}\right).\Gamma\left(\mathrm{4}\right)\:=\frac{\pi^{\mathrm{4}} }{\mathrm{90}}.\mathrm{3}! \\ $$$$=\frac{\mathrm{6}}{\mathrm{90}}.\pi^{\mathrm{4}} \:=\frac{\mathrm{2}.\mathrm{3}}{\mathrm{3}.\mathrm{30}}\pi^{\mathrm{4}} \:=\frac{\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$

Commented by mnjuly1970 last updated on 10/Feb/21

$${thank}\:{you}\:{sir}\:{max}… \\ $$

Commented by mathmax by abdo last updated on 10/Feb/21

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$

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