Answer to Question #102648 in Complex Analysis for BIVEK SAH

Question #102648
Using the method of residues, show that:


Integral( 0 to infinity )(dx/1+x^2)=pie/2
1
Expert's answer
2020-02-20T08:38:39-0500
"\\intop_0^\\infin\\frac{1}{1+x^2}dx=\\frac{\\pi}{2}"


The integrand is even, so the integral can be replaced with


"\\intop_0^\\infin\\frac{1}{1+x^2}dx=\\frac{1}{2}\\intop_{-\\infin}^\\infin\\frac{1}{1+x^2}dx"


We use the formula:


"\\oint_Cf(z)dz=2\\pi i\\sum Res{f(z)}"


Find the singular point(s):


"f(z)=\\frac{1}{1+z^2}=\\frac{1}{(z-i)(z+i)}"

"z=i" is a singular point located in the upper complex half-plane.

"Res_{z=i}f(z)=\\lim_{z\\rightarrow i}f(z)(z-i)=\\lim_{z\\rightarrow i}\\frac{1}{(z+i)(z-i)}(z-i)=\\lim_{z\\rightarrow i}\\frac{1}{z+i}=\\frac{1}{2i}"



"\\intop_{-\\infin}^\\infin\\frac{1}{1+x^2}dx=2\\pi i\\cdot\\frac{1}{2i}=\\pi"

We get


"\\intop_0^\\infin\\frac{1}{1+x^2}dx=\\frac{1}{2}\\intop_{-\\infin}^\\infin\\frac{1}{1+x^2}dx=\\frac{\u03c0}{2}"

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