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Question Number 32740 by caravan msup abdo. last updated on 01/Apr/18

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

Commented by abdo imad last updated on 04/Apr/18

$$letputf\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{ln(x^2+t^2)}{1+t^2}dtwehave$$$$f^{{}^{\prime }}\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{2x}{(x^2+t^2)(1+t^2)}dt=x{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{(x^2+t^2)(1+t^2)}$$$$letuntroducethecomplexfunction$$$$\phi \left(z\right)=\frac{1}{(z^2+x^2)(z^2+1)}=\frac{1}{(z+ix)(z-ix)(z-i)(z+i)}$$$$thepolesof\phi arei,-i,ix,-ix\left(simples\right)$$$$case1x>0$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (Res(\phi ,i)+Res(\phi ,ix))but$$$$Res(\phi ,i)=\frac{1}{2i(i+ix)(i-ix)}=\frac{1}{-2i(1-x^2)}=\frac{i}{2(1-x^2)}$$$$Res(\phi ,ix)=\frac{1}{2ix(ix-i)(ix+i)}=\frac{1}{-2ix(x^2-1)}=\frac{i}{2x(x^2-1)}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (\frac{i}{2(1-x^2)}-\frac{i}{2x(1-x^2)})$$$$=\frac{-\pi }{1-x^2}(1-\frac{1}{x})=-\frac{\pi (x-1)}{x(1-x^2)}=\frac{\pi (1-x)}{x(1-x)(1+x)}=\frac{\pi }{x(1+x)}$$$$f^{{}^{\prime }}\left(x\right)=\frac{\pi }{1+x}withcondition{x}^2\ne 1\Rightarrow$$$$f\left(x\right)=\pi ln\mid 1+x\mid +\lambda$$$$\lambda =f\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{2lnt}{1+t^2}dt=2{\int }_0^{\mathrm{\infty }}\frac{lnt}{1+t^2}dt=0\left(resultproved\right)$$$$\Rightarrow f\left(x\right)=\pi ln\mid 1+x\mid$$$$case2x<0thepolesareiand-ix$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (Res(\phi ,i)+Res(\phi ,-ix))$$$$Res(\phi ,i)=\frac{i}{2(1-x^2)}$$$$Res(\phi ,-ix)=\frac{1}{-2ix(-ix-i)(-ix+i)}$$$$=\frac{1}{-2ix(ix+i)(ix-i)}=\frac{1}{2ix(x^2-1)}=\frac{i}{2x(1-x^2)}$$

Commented by abdo imad last updated on 04/Apr/18

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (\frac{i}{2(1-x^2)}+\frac{i}{2x(1-x^2)})$$$$=-\frac{\pi }{1-x^2}(1+\frac{1}{x})=\frac{-\pi (1+x)}{x(1-x^2)}=\frac{-\pi }{x(1-x)}=\frac{\pi }{x(x-1)}$$$$f^{{}^{\prime }}\left(x\right)=\frac{\pi }{x-1}\Rightarrow f\left(x\right)=\pi ln\mid x-1\mid +\lambda but\lambda =f\left(0\right)=0\Rightarrow$$$$f\left(x\right)=\pi ln\mid 1-x\mid .$$

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