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Question Number 151145 by mnjuly1970 last updated on 18/Aug/21

Answered by qaz last updated on 18/Aug/21

$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{\cosh {}^2\left({\mathrm{x}}^2\right)}\mathrm{dx}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{1+\cosh \left({2\mathrm{x}}^2\right)}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\sqrt{\mathrm{x}}}{1+\cosh \left(2\mathrm{x}\right)}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{2\sqrt{\mathrm{x}}{\mathrm{e}}^{2\mathrm{x}}}{{(1+{\mathrm{e}}^{2\mathrm{x}})}^2}\mathrm{dx}$$$$=-\frac{\sqrt{\mathrm{x}}}{1+{\mathrm{e}}^{2\mathrm{x}}}{\mid }_0^{\mathrm{\infty }}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{-1/2}}{1+{\mathrm{e}}^{2\mathrm{x}}}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{-1/2}{\mathrm{e}}^{-2\mathrm{x}}}{1+{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}$$$$=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{-1/2}{\mathrm{e}}^{-(2\mathrm{n}+2)\mathrm{x}}\mathrm{dx}$$$$=\frac{\sqrt{\pi }}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+2)}^{\frac{1}{2}}}$$$$=\frac{\sqrt{\pi }}{2\sqrt{2}}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{{\mathrm{n}}^{\frac{1}{2}}}$$$$=\frac{\sqrt{\pi }}{2\sqrt{2}}\eta \left(\frac{1}{2}\right)$$$$=\frac{\sqrt{\pi }}{2\sqrt{2}}(1-2^{1-\frac{1}{2}})\zeta \left(\frac{1}{2}\right)$$$$\Rightarrow \mathrm{k}=\frac{\sqrt{\pi }}{2\sqrt{2}}(1-\sqrt{2})$$

Commented by mnjuly1970 last updated on 18/Aug/21

$$verynicemrqaz...$$

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