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Question Number 43939 by abdo.msup.com last updated on 18/Sep/18

$$findf\left(\xi \right)={\int }_0^{\mathrm{\infty }}\frac{dt}{1+{(t-i\xi )}^2}$$$$andcalculate{f}^{{}^{\prime }}\left(\xi \right)$$

Commented by maxmathsup by imad last updated on 22/Sep/18

$$f\left(\xi \right)={\int }_0^{\mathrm{\infty }}\frac{dt}{{(t-\xi )}^2-i^2}={\int }_0^{\mathrm{\infty }}\frac{dt}{((t-\xi )-i)(t-\xi +i)}$$$$=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}\{\frac{1}{t-\xi -i}-\frac{1}{t-\xi +i}\}dt\Rightarrow 2if\left(\xi \right)={\int }_0^{\mathrm{\infty }}\frac{dt}{t-\xi -i}-{\int }_0^{\mathrm{\infty }}\frac{dt}{t-\xi +i}$$$$but\int \frac{dt}{t-\xi -i}=\int \frac{t-\xi +i}{{(t-\xi )}^2+1}dt=\int \frac{t-\xi }{{(t-\xi )}^2+1}dt+i\int \frac{dt}{{(t-\xi )}^2+1}$$$$\int \frac{t-\xi }{{(t-\xi )}^2+1}dt=\frac{1}{2}ln\{{(t-\xi )}^2+1\}+c_1$$$$\int \frac{dt}{{(t-\xi )}^2+1}{=}_{t-\xi =u}\int \frac{du}{1+u^2}=arctan(t-\xi )+c_2\Rightarrow$$$$\int \frac{dt}{t-\xi -i}=\frac{1}{2}ln\{{(t-\xi )}^2+1\}+iarctan(t-\xi )+calso$$$$\int \frac{dt}{t-\xi +i}=\frac{1}{2}ln\{{(t-\xi )}^2+1\}-iarctan(t-\xi )\Rightarrow$$$$2if\left(\xi \right)=2i{\left[arctan(t-\xi )\right]}_0^{+\mathrm{\infty }}=2i\{\frac{\pi }{2}+arctan\left(\xi \right)\}\Rightarrow$$$$f\left(\xi \right)=\frac{\pi }{2}+arctan\left(\xi \right)andwehave$$$$f^{{}^{\prime }}\left(\xi \right)=\frac{1}{1+{\xi }^2}.$$$$$$

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