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Question Number 42504 by maxmathsup by imad last updated on 26/Aug/18

$$letx>0provethat$$ $${\int }_0^{\mathrm{\infty }}\frac{e^{-t^2}ln(1+{xt}^2)}{t^2}dt=\pi {\int }_0^{\sqrt{x}}{e}^{\frac{1}{u^2}}du.$$

Commented bymaxmathsup by imad last updated on 28/Aug/18

$$letf\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-t^2}ln(1+{xt}^2)}{t^2}dt\Rightarrow f^{{}^{\prime }}\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-t^2}}{1+{xt}^2}dt\Rightarrow$$ $$2f^{{}^{\prime }}\left(x\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{-t^2}}{1+{xt}^2}dtletconsiderthecomplexfunction$$ $$\phi \left(z\right)=\frac{e^{-z^2}}{1+{xz}^2}wehave\phi \left(z\right)=\frac{e^{-z^2}}{(\sqrt{x}z-i)(\sqrt{x}z+i)}=\frac{e^{-z^2}}{x(z-\frac{i}{\sqrt{x}})(z+\frac{i}{\sqrt{x})})}$$ $$thepolesof\phi are\stackrel{-}{+}\frac{i}{\sqrt{x}}$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,\frac{i}{\sqrt{x}})butRes(\phi ,\frac{i}{\sqrt{x}})=\frac{e^{\frac{1}{x}}}{x\left(\frac{2i}{\sqrt{x}}\right)}=\frac{e^{\frac{1}{x}}}{2i\sqrt{x}}\Rightarrow$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{\frac{1}{x}}}{2i\sqrt{x}}=\pi \frac{e^{\frac{1}{x}}}{\sqrt{x}}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{\pi }{2}\frac{e^{\frac{1}{x}}}{\sqrt{x}}\Rightarrow$$ $$f\left(x\right)=\frac{\pi }{2}{\int }_0^x\frac{e^{\frac{1}{t}}}{\sqrt{t}}dt+cbutc=f\left(0\right)=0\Rightarrow f\left(x\right)=\frac{\pi }{2}{\int }_0^x\frac{e^{\frac{1}{t}}}{\sqrt{t}}dt$$ $${=}_{\sqrt{t}=u}\frac{\pi }{2}{\int }_0^{\sqrt{x}}\frac{e^{\frac{1}{u^2}}}{u}\left(2u\right)du=\pi {\int }_0^{\sqrt{x}}{e}^{\frac{1}{u^2}}du.$$

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