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Question Number 131866 by mnjuly1970 last updated on 09/Feb/21

$$...advancedcalculus...$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{dx}{x^5(e^{\frac{1}{x}}-1)}=?$$$$$$

Answered by mnjuly1970 last updated on 09/Feb/21

$$solution:$$$$\mathrm{\Omega }\stackrel{x=\frac{1}{t}}{=}{\int }_0^{\mathrm{\infty }}\frac{t^5}{e^t-1}\ast \frac{dt}{t^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{t^3}{e^t-1}dt=\mathrm{\Gamma }\left(4\right)\zeta \left(4\right)=\frac{{\pi }^4}{15}$$$$note:{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}dx=\mathrm{\Gamma }\left(s\right).\zeta \left(s\right)$$

Answered by mathmax by abdo last updated on 10/Feb/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^5({\mathrm{e}}^{\frac{1}{\mathrm{x}}}-1)}\Rightarrow \mathrm{\Phi }{=}_{\frac{1}{\mathrm{x}}=\mathrm{t}}-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^5}{({\mathrm{e}}^{\mathrm{t}}-1)}(-\frac{\mathrm{dt}}{{\mathrm{t}}^2})={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^3}{{\mathrm{e}}^{\mathrm{t}}-1}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}{\mathrm{t}}^3}{1-{\mathrm{e}}^{-\mathrm{t}}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^3{\mathrm{e}}^{-\mathrm{t}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{nt}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^3{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}}\mathrm{dt}{=}_{(\mathrm{n}+1)\mathrm{t}=\mathrm{u}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^3}{{(\mathrm{n}+1)}^3}{\mathrm{e}}^{-\mathrm{u}}\frac{\mathrm{du}}{\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^4}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{4-1}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}=\xi \left(4\right).\mathrm{\Gamma }\left(4\right)=\frac{{\pi }^4}{90}.3!$$$$=\frac{6}{90}.{\pi }^4=\frac{2.3}{3.30}{\pi }^4=\frac{{\pi }^4}{15}$$

Commented by mnjuly1970 last updated on 10/Feb/21

$$thankyousirmax...$$

Commented by mathmax by abdo last updated on 10/Feb/21

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

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