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Question Number 80160 by peter frank last updated on 31/Jan/20

∫_0 ^∞ (x^3 /(e^(2x) −e^x ))

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{\mathrm{2}{x}} −{e}^{{x}} } \\ $$

Commented by mathmax by abdo last updated on 31/Jan/20

I=∫_0 ^∞ (x^3 /(e^(2x) −e^x ))dx ⇒I=∫_0 ^∞ ((e^(−2x) x^3 )/(1−e^(−x) ))dx =∫_0 ^∞ e^(−2x) x^3 (Σ_(n=0) ^∞ e^(−nx) ) =Σ_(n=0) ^∞ ∫_0 ^∞ x^3 e^(−(n+2)x) dx =_((n+2)x=t) Σ_(n=0) ^∞ ∫_0 ^∞ (t^3 /((n+2)^3 )) e^(−t) (dt/((n+2))) =Σ_(n=0) ^∞ (1/((n+2)^4 )) ∫_0 ^∞ t^3 e^(−t) dt let rememer Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt ⇒I =Γ(4)Σ_(n=0) ^∞ (1/((n+2)^4 )) =Γ(4)Σ_(n=2) ^∞ (1/n^4 ) Γ(4)=3!=6 and Σ_(n=2) ^∞ (1/n^4 ) =Σ_(n=1) ^∞ (1/n^4 ) −1 =(π^4 /(90))−1 ⇒ I=6((π^4 /(90))−1) =((3.2)/(3.30))π^4 −6 =(π^4 /(15)) −6

$${I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{3}} }{{e}^{\mathrm{2}{x}} −{e}^{{x}} }{dx}\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\mathrm{2}{x}} \:{x}^{\mathrm{3}} }{\mathrm{1}−{e}^{−{x}} }{dx}\:=\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} {x}^{\mathrm{3}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{3}} \:{e}^{−\left({n}+\mathrm{2}\right){x}} {dx}\:=_{\left({n}+\mathrm{2}\right){x}={t}} \:\:\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{3}} }{\left({n}+\mathrm{2}\right)^{\mathrm{3}} }\:{e}^{−{t}} \frac{{dt}}{\left({n}+\mathrm{2}\right)} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{4}} }\:\int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{3}} \:{e}^{−{t}} \:{dt}\:{let}\:{rememer}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt} \\ $$$$\Rightarrow{I}\:=\Gamma\left(\mathrm{4}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{4}} }\:=\Gamma\left(\mathrm{4}\right)\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\ $$$$\Gamma\left(\mathrm{4}\right)=\mathrm{3}!=\mathrm{6}\:\:\:\:\:{and}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:−\mathrm{1}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{90}}−\mathrm{1}\:\Rightarrow \\ $$$${I}=\mathrm{6}\left(\frac{\pi^{\mathrm{4}} }{\mathrm{90}}−\mathrm{1}\right)\:=\frac{\mathrm{3}.\mathrm{2}}{\mathrm{3}.\mathrm{30}}\pi^{\mathrm{4}} −\mathrm{6}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{15}}\:−\mathrm{6} \\ $$

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