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Question Number 103220 by Dwaipayan Shikari last updated on 13/Jul/20

$${\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x+1}dx$$

Answered by mathmax by abdo last updated on 13/Jul/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^3}{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{x}}}{1+{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{x}}\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{nx}}\right)\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^3{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}\mathrm{dx}{=}_{(\mathrm{n}+1)\mathrm{x}=\mathrm{t}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^3}{{(\mathrm{n}+1)}^3}{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dt}}{(\mathrm{n}+1)}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^4}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^3{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\mathrm{we}\mathrm{know}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\mathrm{\Gamma }\left(\mathrm{x}\right)\Rightarrow$$$$\mathrm{I}=\mathrm{\Gamma }\left(4\right)\times \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{{\mathrm{n}}^4}=-3!\times \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^4}$$$$\mathrm{we}\mathrm{have}\delta \left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^{\mathrm{x}}}=(2^{1-\mathrm{x}}-1)\xi \left(\mathrm{x}\right)\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^4}=\delta \left(4\right)=(2^{-3}-1)\xi \left(4\right)=(\frac{1}{8}-1)\xi \left(4\right)=-\frac{7}{8}\xi \left(4\right)\Rightarrow$$$$\mathrm{I}=-(3!)\times (-\frac{7}{8})\xi \left(4\right)=\frac{3.2.7}{4.2}\xi \left(4\right)=\frac{21}{4}\xi \left(4\right)\mathrm{if}\xi \left(4\right)=\frac{{\pi }^4}{90}\mathrm{we}\mathrm{get}$$$$\mathrm{I}=\frac{21{\pi }^4}{360}=\frac{3.7{\pi }^4}{3.120}\Rightarrow \mathrm{I}=\frac{7{\pi }^4}{120}$$

Commented by Dwaipayan Shikari last updated on 13/Jul/20

$$Thankingyou$$

Commented by abdomsup last updated on 13/Jul/20

$$youarewelcome$$

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