Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convolution proof where f=g=1/(1+x^2)

  1. Apr 27, 2013 #1
    So I am looking for some insight one how I might go about solving this problem.

    I have two equations f and g where [tex]f = g = \frac{1}{(1+x^2)}.[/tex]

    The convolution theorem states that [tex]L(f*g) = L(f)*L(g)[/tex] where L can be either the Laplace transform or the Fourier transform.

    So it will look like this [tex]\mathcal{L} \int_{-\infty}^\infty f(t)g(x-t) dt = \mathcal{L} f(x) \cdot \mathcal{L} g(x)[/tex]

    [tex]\mathcal{L} \int_{-\infty}^\infty \frac{1}{(1+t^2)(t^2-2tx+x^2+1)} dt = 2 \cdot \mathcal{L} \frac{1}{1+x^2} dx[/tex]

    I have to show that the left hand side and the right hand side are the same. I have tried to use both Laplace and Fourier transforms and partial fractions and lots of algebra manipulation as well as looking at integral tables.

    What would be your strategy or the key point you would look at to go about showing this is true?
    Last edited: Apr 27, 2013
  2. jcsd
  3. Apr 27, 2013 #2
    One thing I thought was promising was a polynomial integration with
    [itex]\int_{-\infty}^\infty \frac{1}{t^4 +c_1t^3+c_2t^2+c_3t+c_4} dt [/itex]
    where [itex]c1 = -2x, c2 = 2x^2, c3 = -x, c4 = x^2 + 1 [/itex]
    but I couldn't find anything useful from there.

    I'm interested in seeing this what would your strategies be? What toolbox would you reach for?
  4. Apr 27, 2013 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There shouldn't be a convolution on both sides of your equality. It should be L(f*g) = L(f)L(g) (or vice versa)
  5. Apr 27, 2013 #4
    You are right; I meant to use the * as multiplication on the right side and that was unclear. I will edit it and fix it.
  6. Apr 27, 2013 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Is your main problem evaluating the integral on the left hand side?

    Also the right hand side should be L(f)2 not 2L(f)
  7. Apr 28, 2013 #6
    Yes that is true, I did a terrible job of writing that out.

    My main effort has been evaluating the integral on the left hand side because I believe if I can solve for that or show it equal to the right in some way that is all I need.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook