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Question Number 144269 by alcohol last updated on 24/Jun/21

$${\int }_0^{\mathrm{\infty }}\lfloor \frac{y^3}{e^y-1}\rfloor dy$$

Answered by mathmax by abdo last updated on 24/Jun/21

$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^3}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{x}}}{1-{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{x}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{nx}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^3{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}\mathrm{dx}{=}_{(\mathrm{n}+1)\mathrm{x}=\mathrm{t}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^3}{{(\mathrm{n}+1)}^3}{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dt}}{\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^4}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{4-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\xi \left(4\right).\mathrm{\Gamma }\left(4\right)=\frac{{\pi }^4}{90}\times 3!$$$$=\frac{6}{90}\times {\pi }^4=\frac{2\times 3}{3\times 30}.{\pi }^4=\frac{{\pi }^4}{15}$$

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