Question Number 123720 by mnjuly1970 last updated on 27/Nov/20
$$\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:{calulus}… \\ $$$$\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\Phi=\:\:\int_{\mathrm{0}} ^{\:\:\infty} \frac{{x}^{\mathrm{3}} {e}^{\frac{−{x}}{\mathrm{2}}} }{{sinh}\left(\frac{{x}}{\mathrm{2}}\right)}\:=??? \\ $$
Answered by Dwaipayan Shikari last updated on 27/Nov/20
$$\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} {e}^{−\frac{{x}}{\mathrm{2}}} }{{e}^{\frac{{x}}{\mathrm{2}}} −{e}^{−\frac{{x}}{\mathrm{2}}} }{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{{x}} −\mathrm{1}}{dx} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{3}} {e}^{−{nx}} =\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\int_{\mathrm{0}} ^{\infty} {u}^{\mathrm{3}} {e}^{−{u}} {du}\:\:\:\:\:\:\:{nx}={u} \\ $$$$=\mathrm{2}\Gamma\left(\mathrm{4}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }=\frac{\mathrm{2}\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$
Commented by mnjuly1970 last updated on 27/Nov/20
$${thank}\:{you}\:{mr}\:{Dwaipayan}… \\ $$
Commented by Dwaipayan Shikari last updated on 27/Nov/20
☃️
Answered by mathmax by abdo last updated on 28/Nov/20
$$\Phi\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{3}} \:\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{2}}} }{\mathrm{sh}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\mathrm{dx}\:\mathrm{changement}\:\frac{\mathrm{x}}{\mathrm{2}}=\mathrm{t}\:\mathrm{give} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{8t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{t}} }{\mathrm{sh}\left(\mathrm{t}\right)}\left(\mathrm{2}\right)\mathrm{dt}\:=\mathrm{32}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{t}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }\mathrm{dt} \\ $$$$=\mathrm{32}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{3}} \mathrm{e}^{−\mathrm{2t}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{2t}} }\mathrm{dt}\:=\mathrm{32}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{2t}} \sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{2nt}} \:\mathrm{dt} \\ $$$$=\mathrm{32}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{t}} \mathrm{dt}\:\:=_{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{t}=\mathrm{z}} \:\:\mathrm{32}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{z}^{\mathrm{3}} }{\mathrm{8}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{e}^{−\mathrm{z}} \frac{\mathrm{dz}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{4}} }\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{z}^{\mathrm{3}\:} \mathrm{e}^{−\mathrm{z}} \mathrm{dz}\:=\mathrm{2}\Gamma\left(\mathrm{4}\right)\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{4}} } \\ $$$$=\mathrm{2}\Gamma\left(\mathrm{4}\right)\xi\left(\mathrm{4}\right)\:=\mathrm{2}.\mathrm{3}!\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:=\frac{\mathrm{12}\pi^{\mathrm{4}} }{\mathrm{90}}\:=\frac{\mathrm{6}.\mathrm{2}\pi^{\mathrm{4}} }{\mathrm{6}.\mathrm{15}}\:\Rightarrow\Phi=\frac{\mathrm{2}\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$$$ \\ $$