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Question Number 103220 by Dwaipayan Shikari last updated on 13/Jul/20
∫_0 ^∞ (x^3 /(e^x +1))dx
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$
Answered by mathmax by abdo last updated on 13/Jul/20
I =∫_0 ^∞ (x^3 /(e^x +1))dx ⇒ I =∫_0 ^∞ ((x^3 e^(−x) )/(1+e^(−x) ))dx =∫_0 ^∞ x^(3 ) e^(−x) (Σ_(n=0) ^∞ (−1)^n e^(−nx) )dx =Σ_(n=0) ^∞ (−1)^(n ) ∫_0 ^∞ x^3 e^(−(n+1)x ) dx =_((n+1)x =t) Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ (t^3 /((n+1)^3 )) e^(−t ) (dt/((n+1))) =Σ_(n=0) ^∞ (((−1)^n )/((n+1)^4 )) ∫_0 ^∞ t^3 e^(−t) dt we know ∫_0 ^∞ t^(x−1) e^(−t) dt =Γ(x) ⇒ I =Γ(4)×Σ_(n=1) ^∞ (((−1)^(n−1) )/n^4 ) =−3!×Σ_(n=1) ^∞ (((−1)^n )/n^4 ) we have δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) =(2^(1−x) −1)ξ(x) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^4 ) =δ(4) =(2^(−3) −1)ξ(4) =((1/8)−1)ξ(4) =−(7/8)ξ(4) ⇒ I =−(3!)×(−(7/8))ξ(4) =((3.2.7)/(4.2))ξ(4) =((21)/4)ξ(4) if ξ(4) =(π^4 /(90)) we get I =((21π^4 )/(360)) =((3.7π^4 )/(3.120)) ⇒I =((7π^4 )/(120))
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{e}^{\mathrm{x}} +\mathrm{1}}\mathrm{dx}\:\Rightarrow\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{x}} }\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{3}\:} \mathrm{e}^{−\mathrm{x}} \:\left(\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{nx}} \right)\mathrm{dx} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}\:} \:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{x}^{\mathrm{3}} \:\mathrm{e}^{−\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}\:} \mathrm{dx}\:\:=_{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}\:=\mathrm{t}} \sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{3}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{e}^{−\mathrm{t}\:} \:\frac{\mathrm{dt}}{\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{4}} }\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:\:\mathrm{we}\:\mathrm{know}\:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{x}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\Gamma\left(\mathrm{x}\right)\:\Rightarrow \\ $$$$\mathrm{I}\:=\Gamma\left(\mathrm{4}\right)×\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}^{\mathrm{4}} }\:\:=−\mathrm{3}!×\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{4}} } \\ $$$$\mathrm{we}\:\mathrm{have}\:\delta\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{x}} }\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{x}} −\mathrm{1}\right)\xi\left(\mathrm{x}\right)\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{4}} }\:=\delta\left(\mathrm{4}\right)\:=\left(\mathrm{2}^{−\mathrm{3}} −\mathrm{1}\right)\xi\left(\mathrm{4}\right)\:=\left(\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{1}\right)\xi\left(\mathrm{4}\right)\:=−\frac{\mathrm{7}}{\mathrm{8}}\xi\left(\mathrm{4}\right)\:\Rightarrow \\ $$$$\mathrm{I}\:=−\left(\mathrm{3}!\right)×\left(−\frac{\mathrm{7}}{\mathrm{8}}\right)\xi\left(\mathrm{4}\right)\:=\frac{\mathrm{3}.\mathrm{2}.\mathrm{7}}{\mathrm{4}.\mathrm{2}}\xi\left(\mathrm{4}\right)\:=\frac{\mathrm{21}}{\mathrm{4}}\xi\left(\mathrm{4}\right)\:\:\mathrm{if}\:\xi\left(\mathrm{4}\right)\:=\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{I}\:=\frac{\mathrm{21}\pi^{\mathrm{4}} }{\mathrm{360}}\:=\frac{\mathrm{3}.\mathrm{7}\pi^{\mathrm{4}} }{\mathrm{3}.\mathrm{120}}\:\Rightarrow\mathrm{I}\:=\frac{\mathrm{7}\pi^{\mathrm{4}} }{\mathrm{120}} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Jul/20
Thanking you
$${Thanking}\:{you} \\ $$
Commented by abdomsup last updated on 13/Jul/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$

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