Question Number 43939 by abdo.msup.com last updated on 18/Sep/18
$${find}\:{f}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{1}+\left({t}−{i}\xi\right)^{\mathrm{2}} } \\ $$$${and}\:{calculate}\:{f}^{'} \left(\xi\right) \\ $$
Commented by maxmathsup by imad last updated on 22/Sep/18
![f(ξ) =∫_0 ^∞ (dt/((t−ξ)^2 −i^2 )) =∫_0 ^∞ (dt/(((t−ξ)−i)(t−ξ +i))) =(1/(2i))∫_0 ^∞ { (1/(t−ξ−i)) −(1/(t−ξ +i))}dt ⇒2i f(ξ) =∫_0 ^∞ (dt/(t−ξ−i)) −∫_0 ^∞ (dt/(t−ξ +i)) but ∫ (dt/(t−ξ−i)) =∫ ((t−ξ +i)/((t−ξ)^2 +1)) dt =∫ ((t−ξ)/((t−ξ)^2 +1))dt +i ∫ (dt/((t−ξ)^2 +1)) ∫ ((t−ξ)/((t−ξ)^2 +1))dt =(1/2)ln{ (t−ξ)^2 +1} +c_1 ∫ (dt/((t−ξ)^2 +1)) =_(t−ξ =u) ∫ (du/(1+u^2 )) =arctan(t−ξ) +c_2 ⇒ ∫ (dt/(t−ξ−i)) =(1/2)ln{(t−ξ)^2 +1} +i arctan(t−ξ) +c also ∫ (dt/(t−ξ +i)) = (1/2)ln{(t−ξ)^2 +1}−i arctan(t−ξ) ⇒ 2if(ξ)= 2i [arctan(t−ξ)]_0 ^(+∞) = 2i{ (π/2) +arctan(ξ)} ⇒ f(ξ) =(π/2) +arctan(ξ) and we have f^′ (ξ) = (1/(1+ξ^2 )) .](https://www.tinkutara.com/question/Q44163.png)
$${f}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}−\xi\right)^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\left({t}−\xi\right)−{i}\right)\left({t}−\xi\:+{i}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \:\:\left\{\:\frac{\mathrm{1}}{{t}−\xi−{i}}\:−\frac{\mathrm{1}}{{t}−\xi\:+{i}}\right\}{dt}\:\Rightarrow\mathrm{2}{i}\:{f}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}−\xi−{i}}\:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}−\xi\:+{i}}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${but}\:\int\:\:\frac{{dt}}{{t}−\xi−{i}}\:=\int\:\frac{{t}−\xi\:+{i}}{\left({t}−\xi\right)^{\mathrm{2}} \:+\mathrm{1}}\:{dt}\:=\int\:\frac{{t}−\xi}{\left({t}−\xi\right)^{\mathrm{2}} \:+\mathrm{1}}{dt}\:+{i}\:\int\:\:\frac{{dt}}{\left({t}−\xi\right)^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\int\:\:\:\frac{{t}−\xi}{\left({t}−\xi\right)^{\mathrm{2}} \:+\mathrm{1}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\:\left({t}−\xi\right)^{\mathrm{2}} \:+\mathrm{1}\right\}\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\:\:\frac{{dt}}{\left({t}−\xi\right)^{\mathrm{2}} \:+\mathrm{1}}\:=_{{t}−\xi\:={u}} \:\:\int\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:={arctan}\left({t}−\xi\right)\:+{c}_{\mathrm{2}} \:\:\Rightarrow \\ $$$$\int\:\:\:\:\frac{{dt}}{{t}−\xi−{i}}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\left({t}−\xi\right)^{\mathrm{2}} +\mathrm{1}\right\}\:+{i}\:{arctan}\left({t}−\xi\right)\:+{c}\:{also} \\ $$$$\int\:\:\:\frac{{dt}}{{t}−\xi\:+{i}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\left({t}−\xi\right)^{\mathrm{2}} \:+\mathrm{1}\right\}−{i}\:{arctan}\left({t}−\xi\right)\:\Rightarrow \\ $$$$\mathrm{2}{if}\left(\xi\right)=\:\mathrm{2}{i}\:\left[{arctan}\left({t}−\xi\right)\right]_{\mathrm{0}} ^{+\infty} \:=\:\mathrm{2}{i}\left\{\:\frac{\pi}{\mathrm{2}}\:+{arctan}\left(\xi\right)\right\}\:\Rightarrow \\ $$$${f}\left(\xi\right)\:=\frac{\pi}{\mathrm{2}}\:+{arctan}\left(\xi\right)\:\:\:\:\:{and}\:{we}\:{have}\: \\ $$$${f}^{'} \left(\xi\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+\xi^{\mathrm{2}} }\:. \\ $$$$ \\ $$