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Question Number 86728 by M±th+et£s last updated on 30/Mar/20

$${\int }_0^{\mathrm{\infty }}ln(1+\frac{b^2}{x^2})dx$$

Commented by mathmax by abdo last updated on 30/Mar/20

$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}ln(1+\frac{a}{x^2})dxwitha>0$$$$f^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{1}{x^2(1+\frac{a}{x^2})}dx={\int }_0^{\mathrm{\infty }}\frac{dx}{x^2+a}{=}_{x=\sqrt{a}u}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{a}du}{a(1+u^2)}$$$$=\frac{1}{\sqrt{a}}\times \frac{\pi }{2}=\frac{\pi }{2\sqrt{a}}\Rightarrow f\left(a\right)=\pi \sqrt{a}+C$$$$f\left(0\right)=0=C\Rightarrow f\left(a\right)=\pi \sqrt{a}\Rightarrow {\int }_0^{\mathrm{\infty }}ln(1+\frac{b^2}{x^2})dx=f\left(b^2\right)=\pi \mid b\mid$$$$$$

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