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Question Number 78797 by TawaTawa last updated on 20/Jan/20

$$\mathrm{Show}\mathrm{that}:{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^3}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}=\frac{{\pi }^4}{15}$$

Answered by mind is power last updated on 20/Jan/20

$$={\int }_0^{+\mathrm{\infty }}\frac{{\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{x}}}{1-{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}$$$$\frac{1}{1-{\mathrm{e}}^{-\mathrm{x}}}=\sum _{\mathrm{k}⩾0}{\mathrm{e}}^{-\mathrm{kx}}$$$$\Rightarrow ={\int }_0^{+\mathrm{\infty }}{\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{x}}\sum _{\mathrm{k}⩾0}{\mathrm{e}}^{-\mathrm{kx}}\mathrm{dx}$$$$={\int }_0^{+\mathrm{\infty }}{\mathrm{x}}^3\sum _{\mathrm{k}⩾0}{\mathrm{e}}^{-(1+\mathrm{k})\mathrm{x}}$$$$=\mathrm{\Sigma }{\int }_0^{+\mathrm{\infty }}{\mathrm{x}}^3{\mathrm{e}}^{-(1+\mathrm{k})\mathrm{x}}\mathrm{dx}$$$$\mathrm{u}=(1+\mathrm{k})\mathrm{x}\Rightarrow \mathrm{dx}=\frac{\mathrm{du}}{1+\mathrm{k}}$$$$=\sum _{\mathrm{k}⩾0}{\int }_0^{+\mathrm{\infty }}\frac{{\mathrm{u}}^3}{{(1+\mathrm{k})}^4}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}$$$$\mathrm{one}\mathrm{of}\mathrm{definition}\mathrm{of}\mathrm{\Gamma }\left(\mathrm{x}\right)={\int }_0^{+\mathrm{\infty }}{\mathrm{t}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\sum _{\mathrm{k}⩾0}\frac{\mathrm{\Gamma }\left(4\right)}{{(1+\mathrm{k})}^4}=\sum _{\mathrm{k}⩾0}\frac{6}{{(1+\mathrm{k})}^4}=\sum _{\mathrm{n}⩾1}\frac{6}{{\mathrm{n}}^4}=6\zeta \left(4\right)=6.\frac{{\pi }^4}{90}=\frac{{\pi }^4}{15}$$$${\int }_0^{+\mathrm{\infty }}\frac{{\mathrm{x}}^3}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}=\frac{{\pi }^4}{15}$$$$\mathrm{mor}\mathrm{generaly}\mathrm{if}\mathrm{a}>0$$$${\int }_0^{+\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{a}}}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}=\mathrm{\Gamma }(\mathrm{a}+1).\zeta (\mathrm{a}+1)$$$$$$

Commented by TawaTawa last updated on 20/Jan/20

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by mind is power last updated on 20/Jan/20

$$\mathrm{thanx}\mathrm{sir},\mathrm{most}\mathrm{Welcom}$$

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