Question Number 33341 by prof Abdo imad last updated on 14/Apr/18 | ||
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)} \\ $$ $$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of} \\ $$ $${A}\left(\theta\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\:{x}^{\mathrm{2}} \:+\mathrm{1}\:−{sin}^{\mathrm{2}} \theta\right)} \\ $$ $$\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:. \\ $$ | ||
Commented byprof Abdo imad last updated on 18/Apr/18 | ||
$$\left.\mathrm{1}\right)\:{let}\:{put}\:\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)} \\ $$ $$\mathrm{2}{f}\left({a}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)}\:\:{let}\:{vonsider} \\ $$ $${the}\:{complex}\:{function}\:\varphi\left({z}\right)=\:\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{2}\:} \:+{a}^{\mathrm{2}} \right)} \\ $$ $${the}\:{poles}\:{of}\:\:\varphi\:{is}\:{i},−{i},{ia}\:,−{ia} \\ $$ $${case}\:\mathrm{1}\:\:\:{a}>\mathrm{0}\: \\ $$ $$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{ia}\right)\right)\:{but} \\ $$ $$\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{ia}\right)\left({z}+{ia}\right)}\:\Rightarrow \\ $$ $${Res}\left(\varphi,{i}\right)\:=\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)\left({a}^{\mathrm{2}} \:−\mathrm{1}\right)} \\ $$ $${Res}\left(\varphi,{ia}\right)\:=\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{ia}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\:\:\:\frac{\mathrm{1}}{\left(−\mathrm{2}{i}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:+\frac{\mathrm{1}}{\left(\mathrm{2}{ia}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\right) \\ $$ $$=\:\frac{\mathrm{2}\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\:\:\frac{\mathrm{1}}{\mathrm{2}{a}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\:\frac{\mathrm{1}}{{a}}\:−\mathrm{1}\right) \\ $$ $$=\:\frac{\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\:\frac{\mathrm{1}−{a}}{{a}}\:=\:\frac{\pi}{{a}\left(\mathrm{1}+{a}\right)}\:\:\:\Rightarrow\:{I}\:=\:\:\frac{\pi}{\mathrm{2}{a}\left(\mathrm{1}+{a}\right)} \\ $$ $${case}\mathrm{2}\:\:{a}\:<\mathrm{0}\: \\ $$ $$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,−{ia}\right)\right) \\ $$ $${Res}\left(\varphi,−{ia}\right)\:=\:\:\:\frac{\mathrm{1}}{\left(−\mathrm{2}{ia}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\:\frac{−\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:−\frac{\mathrm{1}}{\mathrm{2}{ia}\left(\:\mathrm{1}−{a}^{\mathrm{2}} \right)}\right) \\ $$ $$=\:\frac{−\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\:\:\mathrm{1}+\:\frac{\mathrm{1}}{{a}}\right)\:=\:\frac{−\pi\left(\mathrm{1}+{a}\right)}{{a}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:=\:\:\frac{−\pi}{{a}\left(\mathrm{1}−{a}\right)} \\ $$ $$=\:\:\frac{\pi}{{a}\left({a}−\mathrm{1}\right)}\:\Rightarrow\:{I}\:\:=\:\:\frac{\pi}{\mathrm{2}{a}\left({a}−\mathrm{1}\right)}\:\:\:{and}\:{we}\:{must} \\ $$ $${study}\:{the}\:{special}\:{case}\:\:{a}\:=\overset{−} {+}\:\mathrm{1}\:. \\ $$ $$ \\ $$ | ||
Commented byprof Abdo imad last updated on 18/Apr/18 | ||
$$\left.\mathrm{2}\right)\:{we}\:{have}\:\:{A}\left(\theta\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\:{x}^{\mathrm{2}} \:+{cos}^{\mathrm{2}} \theta\right)} \\ $$ $$={f}\:\left({cos}\theta\right)\:=\:\:\frac{\pi}{\mathrm{2}\:{cos}\theta\left(\mathrm{1}+{cos}\theta\right)}\:. \\ $$ | ||