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Question Number 33629 by naka3546 last updated on 20/Apr/18
Commented by math khazana by abdo last updated on 22/Apr/18
let put I = ∫_0 ^(π/2) (dx/(1+a^2 tan^2 x)) .changement tanx =t give I = ∫_0 ^∞ (1/(1+a^2 t^2 )) (dt/(1+t^2 )) ⇒ 2I =∫_(−∞) ^(+∞) (dt/((1+t^2 )(1+a^2 t^2 ))) let introduce the complex function ϕ(z) = (1/((1+z^2 )(1+a^2 z^2 ))) let suppose a≠0 we have ϕ(z) = (1/((z−i)(z+i)( az−i)−az+i))) = (1/(a^2 (z−i)(z+i)(z−(i/a))(z +(i/a)))) so the poles of ϕ are i,−i,(i/a),−(i/a) case1 a>0 ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( Res(ϕ,i) +Res(ϕ,(i/a))) Res(ϕ,i) = (1/(2i(1−a^2 ))) Res(ϕ, (i/a)) = (1/(2a^2 (i/a)(1+((i/a))^2 ))) = (1/(2ia(1−(1/a^2 )))) = (a^2 /(2ia(a^2 −1)))=(a/(2i(a^2 −1))) ∫_(−∞) ^(+∞) ϕ(z)dz = 2iπ( (1/(2i(1−a^2 ))) −(a/(2i(1−a^2 )))) = (π/(1−a^2 )) −((2πa)/(1−a^2 )) = (π/(1−a^2 ))(1−a) = (π/(1+a))
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{a}^{\mathrm{2}} {tan}^{\mathrm{2}} {x}}\:\:.{changement} \\ $$$${tanx}\:={t}\:{give}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} }\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{2}{I}\:\:=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} \right)}\:{let}\:{introduce}\:{the} \\ $$$${complex}\:{function}\:\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {z}^{\mathrm{2}} \right)} \\ $$$${let}\:{suppose}\:{a}\neq\mathrm{0}\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left.\left({z}−{i}\right)\left({z}+{i}\right)\left(\:{az}−{i}\right)−{az}+{i}\right)} \\ $$$$=\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}−\frac{{i}}{{a}}\right)\left({z}\:+\frac{{i}}{{a}}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi \\ $$$${are}\:{i},−{i},\frac{{i}}{{a}},−\frac{{i}}{{a}} \\ $$$${case}\mathrm{1}\:\:{a}>\mathrm{0} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,\frac{{i}}{{a}}\right)\right) \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\:\:\frac{\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,\:\frac{{i}}{{a}}\right)\:=\:\:\:\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} \frac{{i}}{{a}}\left(\mathrm{1}+\left(\frac{{i}}{{a}}\right)^{\mathrm{2}} \right)}\:=\:\:\frac{\mathrm{1}}{\mathrm{2}{ia}\left(\mathrm{1}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)}\:=\:\frac{{a}^{\mathrm{2}} }{\mathrm{2}{ia}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{{a}}{\mathrm{2}{i}\left({a}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\left(\:\:\:\frac{\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:−\frac{{a}}{\mathrm{2}{i}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\right) \\ $$$$=\:\frac{\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\:\:−\frac{\mathrm{2}\pi{a}}{\mathrm{1}−{a}^{\mathrm{2}} }\:=\:\frac{\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\mathrm{1}−{a}\right)\:=\:\frac{\pi}{\mathrm{1}+{a}} \\ $$
Commented by math khazana by abdo last updated on 22/Apr/18
2I= (π/(1+a)) ⇒ I =(π/(2(1+a))) case2 a<0 ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( Res(ϕ,i) +Res(ϕ,−(i/a))) Res(ϕ,−(i/a)) = (1/(a^2 (−2(i/a))(1−(1/a^2 )))) = ((a^2 )/(−2ia(a^2 −1))) = (a^2 /(2ia(1−a^2 ))) = (a/(2i(1−a^2 ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( (1/(2i(1−a^2 ))) + (a/(2i(1−a^2 )))) = (π/(1−a^2 )) +((πa)/(1−a^2 )) = (π/(1−a^2 ))(1+a) = (π/(1−a)) ⇒ I = (π/(2(1−a))) but we must study the specials?cases a=+^− 1 and a=0 .
$$\mathrm{2}{I}=\:\frac{\pi}{\mathrm{1}+{a}}\:\Rightarrow\:{I}\:\:=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}\right)} \\ $$$${case}\mathrm{2}\:\:{a}<\mathrm{0} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,−\frac{{i}}{{a}}\right)\right) \\ $$$${Res}\left(\varphi,−\frac{{i}}{{a}}\right)\:=\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(−\mathrm{2}\frac{{i}}{{a}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)} \\ $$$$=\:\:\frac{{a}^{\mathrm{2}} \:}{−\mathrm{2}{ia}\left({a}^{\mathrm{2}} \:−\mathrm{1}\right)}\:=\:\:\frac{{a}^{\mathrm{2}} }{\mathrm{2}{ia}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:=\:\frac{{a}}{\mathrm{2}{i}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\:\:\frac{\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:\:+\:\:\frac{{a}}{\mathrm{2}{i}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\right) \\ $$$$=\:\frac{\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\:\:+\frac{\pi{a}}{\mathrm{1}−{a}^{\mathrm{2}} }\:=\:\frac{\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\mathrm{1}+{a}\right)\:=\:\frac{\pi}{\mathrm{1}−{a}}\:\Rightarrow \\ $$$${I}\:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{a}\right)}\:\:{but}\:{we}\:{must}\:{study}\:{the}\:{specials}?{cases} \\ $$$${a}=\overset{−} {+}\mathrm{1}\:\:{and}\:{a}=\mathrm{0}\:. \\ $$
Answered by sma3l2996 last updated on 21/Apr/18
A=∫_0 ^(π/2) (dx/(1+a^2 tan^2 x)) let t=atanx⇒dx=((adt)/(a^2 +t^2 )) A=a∫_0 ^∞ (dt/((a^2 +t^2 )(1+t^2 ))) (1/((a^2 +t^2 )(1+t^2 )))=((kt+z)/(a^2 +t^2 ))+((rt+d)/(1+t^2 )) with k=r=0 and d=−z=(1/(a^2 −1)) so A=(a/(a^2 −1))∫_0 ^∞ ((1/(1+t^2 ))−(1/(a^2 +t^2 )))dt =(a/(a^2 −1))(∫_0 ^∞ (dt/(1+t^2 ))−∫_0 ^∞ (dt/(a^2 (1+((t/a))^2 )))) =(a/(a^2 −1))[tan^(−1) t]_0 ^∞ −(1/((a^2 −1)))∫_0 ^∞ (1/(1+((t/a))^2 ))d((t/a)) =((aπ)/(2(a^2 −1)))−(π/(2(a^2 −1)))=(π/(2(a^2 −1)))(a−1) A=(π/(2(a+1)))
$${A}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dx}}{\mathrm{1}+{a}^{\mathrm{2}} {tan}^{\mathrm{2}} {x}} \\ $$$${let}\:\:{t}={atanx}\Rightarrow{dx}=\frac{{adt}}{{a}^{\mathrm{2}} +{t}^{\mathrm{2}} } \\ $$$${A}={a}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({a}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{{kt}+{z}}{{a}^{\mathrm{2}} +{t}^{\mathrm{2}} }+\frac{{rt}+{d}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${with}\:\:{k}={r}=\mathrm{0}\:\:{and}\:\:{d}=−{z}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${so}\:\:{A}=\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{t}^{\mathrm{2}} }\right){dt} \\ $$$$=\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}\left(\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{{a}^{\mathrm{2}} \left(\mathrm{1}+\left(\frac{{t}}{{a}}\right)^{\mathrm{2}} \right)}\right) \\ $$$$=\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}\left[{tan}^{−\mathrm{1}} {t}\right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{t}}{{a}}\right)^{\mathrm{2}} }{d}\left(\frac{{t}}{{a}}\right) \\ $$$$=\frac{{a}\pi}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}−\frac{\pi}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{\pi}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\left({a}−\mathrm{1}\right) \\ $$$${A}=\frac{\pi}{\mathrm{2}\left({a}+\mathrm{1}\right)} \\ $$

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