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Question Number 144269 by alcohol last updated on 24/Jun/21

∫_0 ^( ∞) ⌊(y^3 /(e^y −1))⌋dy

$$\int_{\mathrm{0}} ^{\:\infty} \lfloor\frac{{y}^{\mathrm{3}} }{{e}^{{y}} −\mathrm{1}}\rfloor{dy} \\ $$

Answered by mathmax by abdo last updated on 24/Jun/21

∫_0 ^∞ (x^3 /(e^x −1))dx=∫_0 ^∞ ((x^3 e^(−x) )/(1−e^(−x) ))dx =∫_0 ^∞ x^3 e^(−x) Σ_(n=0) ^∞ e^(−nx) dx =Σ_(n=0) ^∞ ∫_0 ^∞ x^3 e^(−(n+1)x) dx =_((n+1)x=t) Σ_(n=0) ^∞ ∫_0 ^∞ (t^3 /((n+1)^3 ))e^(−t) (dt/(n+1)) =Σ_(n=0) ^∞ (1/((n+1)^4 ))∫_0 ^∞ t^(4−1) e^(−t) dt =ξ(4).Γ(4)=(π^4 /(90))×3! =(6/(90))×π^4 =((2×3)/(3×30)).π^4 =(π^4 /(15))

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

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