From Example Problems
Jump to: navigation, search

Evaluate \int _{0}^{\infty }{\frac  {x^{\alpha }}{1+2x\cos \phi +x^{2}}}dx,-1<\alpha <1,\alpha \phi \neq 0,-\pi <\phi <\pi \,.

Factor the bottom.

\int _{0}^{\infty }{\frac  {x^{\alpha }}{(x+e^{{i\phi }})(x+e^{{-i\phi }})}}dx\,

Use a keyhole shaped contour- two concentric circles with radii \epsilon and \rho . Eventually \epsilon \to 0 and \rho \to \infty . The integral around the contour is

\int _{C}dz=\int _{\epsilon }^{\rho }dx+\int _{{C_{\rho }}}dz+\int _{\rho }^{\epsilon }dx+\int _{{C_{\epsilon }}}dz\,

\int _{{C_{\rho }}}\to 0\, by Jordan's Lemma.

\int _{{C_{\epsilon }}}=-2\pi i(0)=0\,

On \int _{\rho }^{\epsilon }\,, x\cong xe^{{i\alpha 2\pi }}\,.

So now,

\int _{\rho }^{\epsilon }=-e^{{i\alpha 2\pi }}\int _{\epsilon }^{\rho }\,

\int _{C}=(1-e^{{i\alpha 2\pi }})\int _{\epsilon }^{\rho }\,

And also from residue theory,

\int _{C}=2\pi i\left[{\mathrm  {Res}}_{{z=-e^{{i\phi }}}}+{\mathrm  {Res}}_{{z=-e^{{-i\phi }}}}\right]\,

{\mathrm  {Res}}_{{z=-e^{{i\phi }}}}={\frac  {e^{{i\alpha (\phi +\pi )}}}{-2i\sin \phi }}\,

{\mathrm  {Res}}_{{z=-e^{{-i\phi }}}}={\frac  {e^{{-i\alpha (\phi +\pi )}}}{2i\sin \phi }}\,


\int _{C}=2\pi i\left[{\frac  {-e^{{i\alpha (\phi +\pi )}}+e^{{-i\alpha (\phi +\pi )}}}{2i\sin \phi }}\right]\,

\lim _{{\epsilon \to 0,\rho \to \infty }}={\frac  {\pi }{\sin \phi }}{\frac  {-e^{{i\alpha (\phi +\pi )}}+e^{{-i\alpha (\phi +\pi )}}}{1-e^{{i\alpha 2\pi }}}}\,

\int _{0}^{\infty }={\frac  {\pi }{\sin \phi }}{\frac  {\sin {\alpha \phi }}{\sin \alpha \pi }}

Main Page : Complex Variables : Residue Calculus