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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

$$letputI={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+a^2{tan}^{2}x}.changement$$$$tanx=tgiveI={\int }_0^{\mathrm{\infty }}\frac{1}{1+a^2{t}^2}\frac{dt}{1+t^2}\Rightarrow$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{(1+t^2)(1+a^2{t}^2)}letintroducethe$$$$complexfunction\phi \left(z\right)=\frac{1}{(1+z^2)(1+a^2{z}^2)}$$$$letsupposea\ne 0wehave$$$$\phi \left(z\right)=\frac{1}{(z-i)(z+i)(az-i)-az+i)}$$$$=\frac{1}{a^2(z-i)(z+i)(z-\frac{i}{a})(z+\frac{i}{a})}sothepolesof\phi$$$$arei,-i,\frac{i}{a},-\frac{i}{a}$$$$case1a>0$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (Res(\phi ,i)+Res(\phi ,\frac{i}{a}))$$$$Res(\phi ,i)=\frac{1}{2i(1-a^2)}$$$$Res(\phi ,\frac{i}{a})=\frac{1}{2a^2\frac{i}{a}(1+{\left(\frac{i}{a}\right)}^2)}=\frac{1}{2ia(1-\frac{1}{a^2})}=\frac{a^2}{2ia(a^2-1)}=\frac{a}{2i(a^2-1)}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (\frac{1}{2i(1-a^2)}-\frac{a}{2i(1-a^2)})$$$$=\frac{\pi }{1-a^2}-\frac{2\pi a}{1-a^2}=\frac{\pi }{1-a^2}(1-a)=\frac{\pi }{1+a}$$

Commented by math khazana by abdo last updated on 22/Apr/18

$$2I=\frac{\pi }{1+a}\Rightarrow I=\frac{\pi }{2(1+a)}$$$$case2a<0$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (Res(\phi ,i)+Res(\phi ,-\frac{i}{a}))$$$$Res(\phi ,-\frac{i}{a})=\frac{1}{a^2(-2\frac{i}{a})(1-\frac{1}{a^2})}$$$$=\frac{a^2}{-2ia(a^2-1)}=\frac{a^2}{2ia(1-a^2)}=\frac{a}{2i(1-a^2)}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (\frac{1}{2i(1-a^2)}+\frac{a}{2i(1-a^2)})$$$$=\frac{\pi }{1-a^2}+\frac{\pi a}{1-a^2}=\frac{\pi }{1-a^2}(1+a)=\frac{\pi }{1-a}\Rightarrow$$$$I=\frac{\pi }{2(1-a)}butwemuststudythespecials?cases$$$$a=\stackrel{-}{+}1anda=0.$$

Answered by sma3l2996 last updated on 21/Apr/18

$$A={\int }_0^{\pi /2}\frac{dx}{1+a^2{tan}^{2}x}$$$$lett=atanx\Rightarrow dx=\frac{adt}{a^2+t^2}$$$$A=a{\int }_0^{\mathrm{\infty }}\frac{dt}{(a^2+t^2)(1+t^2)}$$$$\frac{1}{(a^2+t^2)(1+t^2)}=\frac{kt+z}{a^2+t^2}+\frac{rt+d}{1+t^2}$$$$withk=r=0andd=-z=\frac{1}{a^2-1}$$$$soA=\frac{a}{a^2-1}{\int }_0^{\mathrm{\infty }}(\frac{1}{1+t^2}-\frac{1}{a^2+t^2})dt$$$$=\frac{a}{a^2-1}({\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}-{\int }_0^{\mathrm{\infty }}\frac{dt}{a^2(1+{\left(\frac{t}{a}\right)}^2)})$$$$=\frac{a}{a^2-1}{\left[{tan}^{-1}t\right]}_0^{\mathrm{\infty }}-\frac{1}{(a^2-1)}{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\left(\frac{t}{a}\right)}^2}d\left(\frac{t}{a}\right)$$$$=\frac{a\pi }{2(a^2-1)}-\frac{\pi }{2(a^2-1)}=\frac{\pi }{2(a^2-1)}(a-1)$$$$A=\frac{\pi }{2(a+1)}$$

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