Question Number 80160 by peter frank last updated on 31/Jan/20
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{\mathrm{2}{x}} −{e}^{{x}} } \\ $$
Commented by mathmax by abdo last updated on 31/Jan/20
$${I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{3}} }{{e}^{\mathrm{2}{x}} −{e}^{{x}} }{dx}\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\mathrm{2}{x}} \:{x}^{\mathrm{3}} }{\mathrm{1}−{e}^{−{x}} }{dx}\:=\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} {x}^{\mathrm{3}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{3}} \:{e}^{−\left({n}+\mathrm{2}\right){x}} {dx}\:=_{\left({n}+\mathrm{2}\right){x}={t}} \:\:\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{3}} }{\left({n}+\mathrm{2}\right)^{\mathrm{3}} }\:{e}^{−{t}} \frac{{dt}}{\left({n}+\mathrm{2}\right)} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{4}} }\:\int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{3}} \:{e}^{−{t}} \:{dt}\:{let}\:{rememer}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt} \\ $$$$\Rightarrow{I}\:=\Gamma\left(\mathrm{4}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{4}} }\:=\Gamma\left(\mathrm{4}\right)\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\ $$$$\Gamma\left(\mathrm{4}\right)=\mathrm{3}!=\mathrm{6}\:\:\:\:\:{and}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:−\mathrm{1}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{90}}−\mathrm{1}\:\Rightarrow \\ $$$${I}=\mathrm{6}\left(\frac{\pi^{\mathrm{4}} }{\mathrm{90}}−\mathrm{1}\right)\:=\frac{\mathrm{3}.\mathrm{2}}{\mathrm{3}.\mathrm{30}}\pi^{\mathrm{4}} −\mathrm{6}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{15}}\:−\mathrm{6} \\ $$