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Question Number 172306 by mathocean1 last updated on 25/Jun/22

$$showthatJ={\int }_0^{+\mathrm{\infty }}\frac{ln\left(t\right)}{t^2+a^2}dtwitha>0$$ $$isconvergent$$

Answered by aleks041103 last updated on 25/Jun/22

$$\frac{1}{a^2}{\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)dt}{{\left(t/a\right)}^2+1}=\frac{1}{a^2}{\int }_0^{\mathrm{\infty }}\frac{ln\left(ax\right)d\left(ax\right)}{x^2+1}=$$ $$=\frac{1}{a}[{\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)dx}{1+x^2}+ln\left(a\right){\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^2}]=$$ $$=\frac{\pi ln\left(a\right)}{2a}+\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)dx}{1+x^2}=$$ $$=\frac{\pi ln\left(a\right)}{2a}+\frac{1}{a}I$$ $$I={\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)dx}{1+x^2}={\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+{\left(1/x\right)}^2}\frac{dx}{x^2}=$$ $$={\int }_0^{\mathrm{\infty }}\frac{-ln\left(1/x\right)}{1+{\left(1/x\right)}^2}d(-1/x)={\int }_{\mathrm{\infty }}^0\frac{ln\left(u\right)du}{1+u^2}=-I$$ $$\Rightarrow I=-I\Rightarrow I=0$$ $$\Rightarrow J\left(a\right)=\frac{\pi ln\left(a\right)}{2a}$$ $$Thisisobviouslyfinitefora>0.$$

Answered by Mathspace last updated on 25/Jun/22

$$I{=}_{t=ax}{\int }_0^{\mathrm{\infty }}\frac{lna+lnx}{a^2{x}^2+a^2}adx$$ $$=\frac{lna}{a}{\int }_0^{\mathrm{\infty }}\frac{dx}{x^2+1}+\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{lnx}{x^2+1}dx(\to 0)$$ $$=\frac{lna}{a}\times \frac{\pi }{2}=\frac{\pi lna}{2a}$$

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