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Question Number 32478 by prof Abdo imad last updated on 25/Mar/18

$$find{\int }_0^{\mathrm{\infty }}ln\left(\frac{1+t^2}{t^2}\right)dt$$

Commented by prof Abdo imad last updated on 30/Mar/18

$$I={\int }_0^{+\mathrm{\infty }}ln(1+\frac{1}{t^2})dtbyparts$$$$I=]{\left[tln(1+\frac{1}{t^2})\right]}_0^{+\mathrm{\infty }}-{\int }_0^{+\mathrm{\infty }}t\frac{-2}{t^3}{(1+\frac{1}{t^2})}^{-1}dt$$$$=2{\int }_0^{\mathrm{\infty }}\frac{1}{t^2}.\frac{1}{1+\frac{1}{t^2}}dt=2{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+1}=2\frac{\pi }{2}=\pi$$$$letprovethat{lim}_{t\to +\mathrm{\infty }}tln(1+\frac{1}{t^2})=0$$$$={lim}_{u\to 0}\frac{1}{u}ln(1+u^2)={lim}_{u\to \mathrm{\infty }}u\frac{ln(1+u^2)}{u^2}=0$$$$also{lim}_{t\to o}tln(1+\frac{1}{t^2})={lim}_{t\to 0}tln(1+t^2)-2tlnt$$$$=0finallyI=\pi$$$$$$

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