Results 1 to 3 of 3

Math Help - Complex Analysis, Integration

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    147

    Complex Analysis, Integration

    Derive \int_0^\infty \frac {(x^{a-1})} {(1+x^2)} dx= \frac {\pi} {(2\sin(\frac {\pi a} {2}))}, 0< a < 2

    Evaluate \int_0^{2\pi} \frac {\cos \theta} {a-\cos \theta} d\theta, a>1

    Evaluate \frac {1} {2\pi i} \int_C \frac {z^{n-1}} {3z^n -1} dz where n is a positive integer, and C is the unit circle with the coutnerclockwise orientation.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by terr13 View Post
    Derive \int_0^\infty \frac {(x^{a-1})} {(1+x^2)} dx= \frac {\pi} {(2\sin(\frac {\pi a} {2}))}, 0< a < 2
    I think you mean 0 < a < 1 anyway if you use keyhole contour you get:

    \left( 1 - e^{2\pi i (a-1)} \right) \int_0^{\infty} \frac{x^{a-1}}{x^2+1}dx = 2\pi i \left( \text{res}(f,i) + \text{res}(f,-i) \right) where f(z) = \frac{z^{a-1}}{z^2+1}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by terr13 View Post
    Evaluate \int_0^{2\pi} \frac {\cos \theta} {a-\cos \theta} d\theta, a>1
    Notice that \int_0^{2\pi} \frac{\cos \theta}{a - \cos \theta} d\theta = \oint_{|z|=1} \frac{\frac{1}{2}(z+z^{-1})}{a - \frac{1}{2}(z+z^{-1})}\cdot \frac{1}{iz} dz.

    Simplify, \frac{1}{i}\oint_{|z|=1} \frac{z^2+1}{z(2az - z^2 - 1)} dz

    Let f(z) = \frac{z^2+1}{z(2az - z^2 - 1)} then the poles are when z(2az-z^2-1) = 0. One is easy z=0 the other is when z^2 - 2az + 1 = 0 and that happens for z_{1,2}= a \pm \sqrt{a^2-1}. Since we require |z|<1 (we are inside the disk) it means z_1 = a - \sqrt{a^2-1}.

    Now f(z) = \frac{g(z)}{h(z)} with h'(z_1)\not = 0 so \text{res}(f,z_1) = \frac{g(z_1)}{h'(z_1)} =\frac{z_1^2 + 1}{(2az_1-z_1^2-1) + z_1(2a - 2z_1)}.

    Thus, \text{res}(f,z_1) = \frac{2az_1}{2z_1(a-z_1)} = \frac{a}{a-z_1} = \frac{a}{\sqrt{a^2-1}}.

    While, \text{res}(f,0) = \frac{g(0)}{h'(0)} = -1.

    Thus, the integral is 2\pi \left( \frac{a}{\sqrt{a^2-1}} - 1 \right)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with integration (complex analysis)
    Posted in the Calculus Forum
    Replies: 10
    Last Post: October 28th 2009, 10:58 AM
  2. complex analysis - integration
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: June 8th 2009, 12:44 PM
  3. complex analysis integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 5th 2008, 09:48 PM
  4. Complex Analysis - Integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 4th 2007, 08:51 PM
  5. Complex Analysis Integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 16th 2007, 11:19 AM

Search Tags


/mathhelpforum @mathhelpforum