Question Number 31083 by abdo imad last updated on 02/Mar/18
$${calculate}\:{by}\:{two}\:{methods}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}\:{dt}}{\mathrm{1}+{x}^{\mathrm{2}} {tan}^{\mathrm{2}} {t}} \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{t}\:{cotant}\:{dt}\:. \\ $$$$ \\ $$
Commented by abdo imad last updated on 11/Mar/18
![let put I= ∫_0 ^(π/2) (∫_0 ^1 (dx/(1+x^2 tan^2 t)))dt ch. xtant=u give ∫_0 ^1 (dx/(1+x^2 tan^2 t))= ∫_0 ^1 (1/(1+u^2 )) (du/(tant)) =(1/(tant)) [arctanu]_0 ^(tant) = (t/(tant)) ⇒ I= ∫_0 ^(π/2) t cotant dt from another side by fubini theorem we have I= ∫_0 ^1 (∫_0 ^(π/2) (dt/(1+x^2 tan^2 t)))dx the ch.tant =u give ∫_0 ^(π/2) (dt/(1+x^2 tan^2 t)) = ∫_0 ^∞ (du/((1+u^2 )(1+x^2 u^2 ))) let decompose F(u)= (1/((1+u^2 )(1+x^2 u^2 ))) =((au+b)/(1+u^2 )) +((cu +d)/(1+x^2 u^2 )) F(−u)=F(u) ⇒((−au +b)/(1+u^2 )) +((−cu+d)/(1+x^2 u^2 ))=F(u) ⇒a=c=0⇒ F(u) =(b/(1+u^2 )) +(d/(1+x^2 u^2 )) lim_(u→∞) u^2 F(u)=0=b +(d/x^2 ) ⇒ d=−bx^2 ⇒F(u)= (b/(1+u^2 )) −((bx^2 )/(1+x^2 u^2 )) we look that b=(1/(1−x^2 )) and F(u)=(1/(1−x^2 ))( (1/(1+u^2 )) −(x^2 /(1+x^2 u^2 ))) ∫_0 ^∞ F(u)du= (1/(1−x^2 ))∫_0 ^∞ (du/(1+u^2 )) −(x^2 /(1−x^2 )) ∫_0 ^(+∞) (du/(1+x^2 u^2 )) =(π/(2(1−x^2 ))) −(x^2 /(1−x^2 )) ∫_0 ^∞ (1/(1+α^2 ))(dα/x) (ch.xu=α) = (π/(2(1−x^2 ))) −((πx)/(2(1−x^2 )))=((π(1−x))/(2(1−x)(1+x))) = (π/(2(1+x))) I =∫_0 ^1 ((πdx)/(2(1+x))) =(π/2) ∫_0 ^1 (dx/(1+x)) =(π/2) ln(2) ⇒ ∫_0 ^(π/2) t cotant dt =(π/2)ln(2).](https://www.tinkutara.com/question/Q31627.png)
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} {tan}^{\mathrm{2}} {t}}\right){dt}\:\:{ch}.\:{xtant}={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} {tan}^{\mathrm{2}} {t}}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{{tant}} \\ $$$$=\frac{\mathrm{1}}{{tant}}\:\left[{arctanu}\right]_{\mathrm{0}} ^{{tant}} =\:\frac{{t}}{{tant}}\:\Rightarrow\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {t}\:{cotant}\:{dt} \\ $$$${from}\:{another}\:{side}\:{by}\:{fubini}\:{theorem}\:{we}\:{have} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{1}+{x}^{\mathrm{2}} {tan}^{\mathrm{2}} {t}}\right){dx}\:\:\:{the}\:{ch}.{tant}\:={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{1}+{x}^{\mathrm{2}} {tan}^{\mathrm{2}} {t}}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}\:{let} \\ $$$${decompose}\:{F}\left({u}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}\:=\frac{{au}+{b}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{{cu}\:+{d}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} } \\ $$$${F}\left(−{u}\right)={F}\left({u}\right)\:\Rightarrow\frac{−{au}\:+{b}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{−{cu}+{d}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }={F}\left({u}\right)\:\Rightarrow{a}={c}=\mathrm{0}\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{b}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{{d}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:{lim}_{{u}\rightarrow\infty} {u}^{\mathrm{2}} {F}\left({u}\right)=\mathrm{0}={b}\:+\frac{{d}}{{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${d}=−{bx}^{\mathrm{2}} \:\:\:\Rightarrow{F}\left({u}\right)=\:\frac{{b}}{\mathrm{1}+{u}^{\mathrm{2}} }\:−\frac{{bx}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:{we}\:{look}\:{that}\:{b}=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${and}\:{F}\left({u}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\left(\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:−\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{F}\left({u}\right){du}=\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:−\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{du}}{\mathrm{1}+{x}^{\mathrm{2}} \:{u}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\frac{{d}\alpha}{{x}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({ch}.{xu}=\alpha\right) \\ $$$$=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:−\frac{\pi{x}}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}=\frac{\pi\left(\mathrm{1}−{x}\right)}{\mathrm{2}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}\:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{x}\right)} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\pi{dx}}{\mathrm{2}\left(\mathrm{1}+{x}\right)}\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}}\:=\frac{\pi}{\mathrm{2}}\:{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{t}\:{cotant}\:{dt}\:=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\ $$