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Calculate real integrals using complex analysis

  1. Apr 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate real integrals using complex analysis
    a) ##\int_{-\infty}^{\infty}\frac{dx}{x^2+1}##
    b) ##\int_0^\infty \frac{sin(x)}{x}dx##


    2. Relevant equations

    3. The attempt at a solution

    a)
    ##\int_{-\infty }^{\infty }\frac{dz}{z^2+1}=\int_{-R}^{R}\frac{dx}{x^2+1}+\int _\gamma\frac{dz}{z^2+1}##

    Where ##R-> \infty##.

    If ##z=re^{i\varphi }## than

    ##\int _\gamma\frac{dz}{z^2+1}=i\int _\gamma\lim_{R->\infty }\frac{re^{i\varphi }}{r^2e^{2i\varphi }+1}d\varphi =0##

    So initial equation ##\int_{-\infty }^{\infty }\frac{dz}{z^2+1}=\int_{-R}^{R}\frac{dx}{x^2+1}+\int _\gamma\frac{dz}{z^2+1}## is now ##\int_{-\infty }^{\infty }\frac{dz}{z^2+1}=\int_{-R}^{R}\frac{dx}{x^2+1}##.

    There is a pole of first order in ##i##. Which gives me ##Res(f,i)=\frac{1}{2i}## and finally

    ##\int_{-\infty }^{\infty }\frac{dz}{z^2+1}=2\pi i\frac{1}{2i}=\pi ##

    Therefore ##\int_{-\infty }^{\infty }\frac{dx}{x^2+1}=\pi ##.

    b) Have no idea.

    ##\int_{0}^{\infty}\frac{sin(z)}{z}dz=\int_{0}^{\infty}\frac{sin(x)}{x}dx+\int_\gamma \frac{sin(z)}{z}dz##

    I tried to integrate (because it is also from ##0## to ##\infty ##) ##\int_{0}^{\infty}\frac{dx}{1+x^3}## to maybe find out anything yet I could get the right result here either...

    Lets get back to this: ##\int_{0}^{\infty}\frac{sin(z)}{z}dz##.

    This integral is ##0##, because ##\frac{sin(z)}{z}## has a removable singularity for ##z=0##, therefore ##Res(f,z=0)=0##.

    But what to do with ##\int_\gamma \frac{sin(z)}{z}dz## ... That I do not know.
     
  2. jcsd
  3. Apr 6, 2014 #2

    vela

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    Try using the fact that the integrand is even and that sin x = Im(eix).
     
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