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Question Number 33883 by math khazana by abdo last updated on 26/Apr/18

find a simple form of f(x)=∫_0 ^(π/2) ln(1+xsin^2 t)dt with ∣x∣<1.

$${find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xsin}^{\mathrm{2}} {t}\right){dt} \\ $$ $${with}\:\mid{x}\mid<\mathrm{1}. \\ $$

Commented bymath khazana by abdo last updated on 01/May/18

we have f^′ (x)= ∫_0 ^(π/2) ((sin^2 t)/(1+xsin^2 t))dt and for x≠0 f^′ (x) =(1/x) ∫_0 ^(π/2) ((1+xsin^2 t−1)/(1+xsin^2 t))dt = (π/(2x)) −(1/x) ∫_0 ^(π/2) (dt/(1+x sin^2 t)) but ∫_0 ^(π/2) (dt/(1+x sin^2 t)) = ∫_0 ^(π/2) (dt/(1+x ((1−cos(2t))/2))) = ∫_0 ^(π/2) ((2dt)/(2 +x −x cos(2t))) =_(2t=u) ∫_0 ^π (du/(2+x −x cosu)) also ch tan((u/2))=t give ∫_0 ^(π/2) (dt/(1+x sin^2 t)) = ∫_0 ^(+∞) (1/(2+x−x((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^∞ ((2dt)/((2+x)(1+t^2 ) −x(1−t^2 ))) =∫_0 ^∞ ((2dt)/(2+x +(2+x)t^2 −x +xt^2 )) = ∫_0 ^∞ ((2dt)/(2 + 2(1+x)t^2 )) = ∫_0 ^∞ (dt/(1+(1+x)t^2 )) =_((√(1+x))t=u) ∫_0 ^∞ (1/(1+u^2 )) (du/(√(1+x))) = (π/(2(√(1+x)))) ⇒ f^′ (x) = (π/(2x)) −(1/x) (π/(2(√(1+x)))) = (π/(2x))( 1−(1/(√(1+x)))) ⇒ f(x)=(π/2)ln∣x∣ −(π/2)∫_. ^x (dt/(t(√(1+t)))) +λ ∫ (dt/(t(√(1+t)))) =_((√(1+t))=u) ∫ ((2udu)/((u^2 −1)u)) = ∫ ((2du)/((u+1)(u−1))) =∫ ((1/(u−1)) −(1/(u+1)))du =ln∣((u−1)/(u+1))∣ =ln((((√(1+x))−1)/((√(1+x)) +1))) ⇒ f(x) = (π/2)ln∣x∣ −(π/2)ln((((√(1+x))−1)/((√(1+x)) +1))) +λ λ =lim_(x→1) (f(x) −(π/2)ln∣x∣ +(π/2)ln((((√(1+x))−1)/((√(1+x))+1)))) =f(1) +(π/2)ln((((√2)−1)/((√2)+1))) =∫_0 ^(π/2) ln(1+sin^2 t)dt +(π/2)ln((((√2)−1)/((√2)+1))) .

$${we}\:{have}\:{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{sin}^{\mathrm{2}} {t}}{\mathrm{1}+{xsin}^{\mathrm{2}} {t}}{dt}\:\:{and}\:{for}\:{x}\neq\mathrm{0} \\ $$ $${f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{xsin}^{\mathrm{2}} {t}−\mathrm{1}}{\mathrm{1}+{xsin}^{\mathrm{2}} {t}}{dt} \\ $$ $$=\:\frac{\pi}{\mathrm{2}{x}}\:\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{x}\:{sin}^{\mathrm{2}} {t}}\:\:{but}\: \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{x}\:{sin}^{\mathrm{2}} {t}}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{x}\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}} \\ $$ $$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}\:+{x}\:−{x}\:{cos}\left(\mathrm{2}{t}\right)}\:=_{\mathrm{2}{t}={u}} \:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\:\frac{{du}}{\mathrm{2}+{x}\:−{x}\:{cosu}} \\ $$ $${also}\:{ch}\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)={t}\:{give} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{x}\:{sin}^{\mathrm{2}} {t}}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}+{x}−{x}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$ $$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{2}+{x}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:−{x}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$ $$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+{x}\:\:+\left(\mathrm{2}+{x}\right){t}^{\mathrm{2}} \:\:−{x}\:\:+{xt}^{\mathrm{2}} } \\ $$ $$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}\:\:+\:\mathrm{2}\left(\mathrm{1}+{x}\right){t}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dt}}{\mathrm{1}+\left(\mathrm{1}+{x}\right){t}^{\mathrm{2}} } \\ $$ $$=_{\sqrt{\mathrm{1}+{x}}{t}={u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\sqrt{\mathrm{1}+{x}}}\:=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{1}+{x}}}\:\:\Rightarrow \\ $$ $${f}^{'} \left({x}\right)\:=\:\frac{\pi}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{{x}}\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{1}+{x}}}\:=\:\frac{\pi}{\mathrm{2}{x}}\left(\:\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}}}\right)\:\Rightarrow \\ $$ $${f}\left({x}\right)=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:−\frac{\pi}{\mathrm{2}}\int_{.} ^{{x}} \:\:\:\:\frac{{dt}}{{t}\sqrt{\mathrm{1}+{t}}}\:\:+\lambda \\ $$ $$\int\:\:\:\:\frac{{dt}}{{t}\sqrt{\mathrm{1}+{t}}}\:=_{\sqrt{\mathrm{1}+{t}}={u}} \int\:\:\:\:\frac{\mathrm{2}{udu}}{\left({u}^{\mathrm{2}} −\mathrm{1}\right){u}} \\ $$ $$=\:\int\:\:\:\:\:\frac{\mathrm{2}{du}}{\left({u}+\mathrm{1}\right)\left({u}−\mathrm{1}\right)}\:=\int\:\left(\frac{\mathrm{1}}{{u}−\mathrm{1}}\:−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du} \\ $$ $$={ln}\mid\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\mid\:={ln}\left(\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{1}}{\sqrt{\mathrm{1}+{x}}\:+\mathrm{1}}\right)\:\Rightarrow \\ $$ $${f}\left({x}\right)\:=\:\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:−\frac{\pi}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{1}}{\sqrt{\mathrm{1}+{x}}\:+\mathrm{1}}\right)\:+\lambda \\ $$ $$\lambda\:={lim}_{{x}\rightarrow\mathrm{1}} \left({f}\left({x}\right)\:−\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:+\frac{\pi}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{1}}{\sqrt{\mathrm{1}+{x}}+\mathrm{1}}\right)\right) \\ $$ $$={f}\left(\mathrm{1}\right)\:+\frac{\pi}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{2}}+\mathrm{1}}\right) \\ $$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\mathrm{1}+{sin}^{\mathrm{2}} {t}\right){dt}\:+\frac{\pi}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{2}}+\mathrm{1}}\right)\:. \\ $$ $$ \\ $$

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