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Question Number 123720 by mnjuly1970 last updated on 27/Nov/20

... nice calulus... evaluate :: Φ= ∫_0 ^( ∞) ((x^3 e^((−x)/2) )/(sinh((x/2)))) =???

$$\:\:\:\:\:\:\:\:\:\:...\:{nice}\:\:\:{calulus}... \\ $$$$\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\Phi=\:\:\int_{\mathrm{0}} ^{\:\:\infty} \frac{{x}^{\mathrm{3}} {e}^{\frac{−{x}}{\mathrm{2}}} }{{sinh}\left(\frac{{x}}{\mathrm{2}}\right)}\:=??? \\ $$

Answered by Dwaipayan Shikari last updated on 27/Nov/20

2∫_0 ^∞ ((x^3 e^(−(x/2)) )/(e^(x/2) −e^(−(x/2)) ))dx =2∫_0 ^∞ (x^3 /(e^x −1))dx =2Σ_(n=1) ^∞ ∫_0 ^∞ x^3 e^(−nx) =2Σ_(n=1) ^∞ (1/n^4 )∫_0 ^∞ u^3 e^(−u) du nx=u =2Γ(4)Σ_(n=1) ^∞ (1/n^4 )=((2π^4 )/(15))

$$\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} {e}^{−\frac{{x}}{\mathrm{2}}} }{{e}^{\frac{{x}}{\mathrm{2}}} −{e}^{−\frac{{x}}{\mathrm{2}}} }{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{{x}} −\mathrm{1}}{dx} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{3}} {e}^{−{nx}} =\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\int_{\mathrm{0}} ^{\infty} {u}^{\mathrm{3}} {e}^{−{u}} {du}\:\:\:\:\:\:\:{nx}={u} \\ $$$$=\mathrm{2}\Gamma\left(\mathrm{4}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }=\frac{\mathrm{2}\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$

Commented by mnjuly1970 last updated on 27/Nov/20

thank you mr Dwaipayan...

$${thank}\:{you}\:{mr}\:{Dwaipayan}... \\ $$

Commented by Dwaipayan Shikari last updated on 27/Nov/20

�� ����☃️

Answered by mathmax by abdo last updated on 28/Nov/20

Φ =∫_0 ^∞ ((x^3 e^(−(x/2)) )/(sh((x/2))))dx changement (x/2)=t give Φ=∫_0 ^∞ ((8t^3 e^(−t) )/(sh(t)))(2)dt =32 ∫_0 ^∞ ((t^3 e^(−t) )/(e^t −e^(−t) ))dt =32 ∫_0 ^∞ ((t^3 e^(−2t) )/(1−e^(−2t) ))dt =32 ∫_0 ^∞ t^3 e^(−2t) Σ_(n=0) ^∞ e^(−2nt) dt =32 Σ_(n=0) ^∞ ∫_0 ^∞ t^3 e^(−2(n+1)t) dt =_(2(n+1)t=z) 32Σ_(n=0) ^∞ ∫_0 ^∞ (z^3 /(8(n+1)^3 ))e^(−z) (dz/(2(n+1))) =2 Σ_(n=0) ^∞ (1/((n+1)^4 )) ∫_0 ^∞ z^(3 ) e^(−z) dz =2Γ(4)Σ_(n=1) ^∞ (1/n^4 ) =2Γ(4)ξ(4) =2.3!(π^4 /(90)) =((12π^4 )/(90)) =((6.2π^4 )/(6.15)) ⇒Φ=((2π^4 )/(15))

$$\Phi\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{3}} \:\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{2}}} }{\mathrm{sh}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\mathrm{dx}\:\mathrm{changement}\:\frac{\mathrm{x}}{\mathrm{2}}=\mathrm{t}\:\mathrm{give} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{8t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{t}} }{\mathrm{sh}\left(\mathrm{t}\right)}\left(\mathrm{2}\right)\mathrm{dt}\:=\mathrm{32}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{t}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }\mathrm{dt} \\ $$$$=\mathrm{32}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{3}} \mathrm{e}^{−\mathrm{2t}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{2t}} }\mathrm{dt}\:=\mathrm{32}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{2t}} \sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{2nt}} \:\mathrm{dt} \\ $$$$=\mathrm{32}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{t}} \mathrm{dt}\:\:=_{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{t}=\mathrm{z}} \:\:\mathrm{32}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{z}^{\mathrm{3}} }{\mathrm{8}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{e}^{−\mathrm{z}} \frac{\mathrm{dz}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{4}} }\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{z}^{\mathrm{3}\:} \mathrm{e}^{−\mathrm{z}} \mathrm{dz}\:=\mathrm{2}\Gamma\left(\mathrm{4}\right)\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{4}} } \\ $$$$=\mathrm{2}\Gamma\left(\mathrm{4}\right)\xi\left(\mathrm{4}\right)\:=\mathrm{2}.\mathrm{3}!\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:=\frac{\mathrm{12}\pi^{\mathrm{4}} }{\mathrm{90}}\:=\frac{\mathrm{6}.\mathrm{2}\pi^{\mathrm{4}} }{\mathrm{6}.\mathrm{15}}\:\Rightarrow\Phi=\frac{\mathrm{2}\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$$$ \\ $$

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