Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com