Question Number 32705 by caravan msup abdo. last updated on 31/Mar/18
$${let}\:{give}\:\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:{ln}\left(\mathrm{1}\:+\frac{{x}}{{t}^{\mathrm{2}} }\right){dt}\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$${find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right). \\ $$
Commented by abdo imad last updated on 03/Apr/18
![f^′ (x) =∫_0 ^∞ (1/(t^2 (1+(x/t^2 ))))dt =∫_0 ^∞ (dt/(x +t^2 )) case1 0<x<1 f^′ (x) =_(t=(√x) u) ∫_0 ^∞ (1/(x(1+t^2 ))) (√x) du = (π/(2(√x))) ⇒ f(x)= π(√x) +λ but λ =f(0)=0 ⇒f(x)=π(√x) −1<x<0 ⇒ f^′ (x) =∫_0 ^∞ (dt/(t^2 −((√(−x)))^2 )) = (1/(2(√(−x)))) ∫_0 ^∞ ( (1/(t−(√x))) −(1/(t +(√(−x)))))dt = (1/(2(√(−x)))) [ ln∣ ((t−(√(−x)))/(t+(√(−x)))) ∣]_0 ^∞ =0](https://www.tinkutara.com/question/Q32857.png)
$${f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left(\mathrm{1}+\frac{{x}}{{t}^{\mathrm{2}} }\right)}{dt}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{{x}\:+{t}^{\mathrm{2}} } \\ $$$${case}\mathrm{1}\:\:\:\mathrm{0}<{x}<\mathrm{1}\:\:{f}^{'} \left({x}\right)\:=_{{t}=\sqrt{{x}}\:{u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\sqrt{{x}}\:{du} \\ $$$$=\:\frac{\pi}{\mathrm{2}\sqrt{{x}}}\:\:\Rightarrow\:{f}\left({x}\right)=\:\pi\sqrt{{x}}\:+\lambda\:\:{but}\:\lambda\:={f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{f}\left({x}\right)=\pi\sqrt{{x}} \\ $$$$−\mathrm{1}<{x}<\mathrm{0}\:\:\Rightarrow\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−\left(\sqrt{−{x}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}}\:\int_{\mathrm{0}} ^{\infty} \:\left(\:\:\frac{\mathrm{1}}{{t}−\sqrt{{x}}}\:−\frac{\mathrm{1}}{{t}\:+\sqrt{−{x}}}\right){dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}}\:\left[\:{ln}\mid\:\frac{{t}−\sqrt{−{x}}}{{t}+\sqrt{−{x}}}\:\mid\right]_{\mathrm{0}} ^{\infty} \:=\mathrm{0} \\ $$