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Question Number 200549 by sonukgindia last updated on 20/Nov/23

Answered by witcher3 last updated on 20/Nov/23

$${\mathrm{I}}_{12}=0;\mathrm{x}\to \frac{1}{\mathrm{x}}$$$${\mathrm{I}}_{11}={\int }_0^1\frac{{\ln }^2\left(\mathrm{x}\right)}{1+{\mathrm{x}}^2}+{\int }_1^{\mathrm{\infty }}\frac{{\ln }^2\left(\mathrm{x}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}\mathrm{x}\to \frac{1}{\mathrm{x}}$$$$=2{\int }_0^1\frac{{\ln }^2\left(\mathrm{z}\right)}{1+{\mathrm{z}}^2}\mathrm{dz}=2{\int }_0^1{\ln }^2\left(\mathrm{z}\right)(\sum _{\mathrm{n}⩾1}{(-1)}^{\mathrm{n}}{\mathrm{z}}^{2\mathrm{n}}$$$$\mathrm{libneiz}\mathrm{theorem}\mathrm{\Sigma }{(-1)}^{\mathrm{n}}{\mathrm{z}}^{2\mathrm{n}}\mathrm{cv}\mathrm{normaly}\mathrm{in}[0,\mathrm{a}]$$$$\mathrm{a}<1$$$$\Rightarrow {\mathrm{I}}_{12}=2\sum _{\mathrm{n}⩾0}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{z}}^{2\mathrm{n}}{\ln }^2\left(\mathrm{z}\right)$$$$=4\sum _{\mathrm{n}⩾1}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^3}=$$$$=4\frac{{\pi }^3}{32}=\frac{{\pi }^3}{8}$$

Commented by sonukgindia last updated on 20/Nov/23

$$nice$$

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