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Question Number 157961 by mnjuly1970 last updated on 30/Oct/21

$$$$$$provethat:$$$$$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{x}^{2}tanh\left(x\right).e^{-x}dx=\frac{{\pi }^3}{8}-2$$$$$$

Answered by qaz last updated on 30/Oct/21

$${\int }_0^{\mathrm{\infty }}{\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{x}}\tanh \mathrm{xdx}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^2\frac{{\mathrm{e}}^{-\mathrm{x}}-{\mathrm{e}}^{-3\mathrm{x}}}{1+{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^2({\mathrm{e}}^{-\mathrm{x}}-{\mathrm{e}}^{-3\mathrm{x}}){\mathrm{e}}^{-2\mathrm{nx}}\mathrm{dx}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}(\frac{2}{{(2\mathrm{n}+1)}^3}-\frac{2}{{(2\mathrm{n}+3)}^3})$$$$=2\beta \left(3\right)-2(-\beta \left(3\right)+1)$$$$=\frac{{\pi }^3}{8}-2$$

Commented by mnjuly1970 last updated on 30/Oct/21

$$verynicdsirqaz$$

Answered by mnjuly1970 last updated on 30/Oct/21

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