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

1. Apr 27, 2013

### polpol

So I am looking for some insight one how I might go about solving this problem.

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

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

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

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

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. Apr 27, 2013

### polpol

One thing I thought was promising was a polynomial integration with
$\int_{-\infty}^\infty \frac{1}{t^4 +c_1t^3+c_2t^2+c_3t+c_4} dt$
where $c1 = -2x, c2 = 2x^2, c3 = -x, c4 = x^2 + 1$
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?

3. Apr 27, 2013

### Office_Shredder

Staff Emeritus
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)

4. Apr 27, 2013

### polpol

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.

5. Apr 27, 2013

### Office_Shredder

Staff Emeritus
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)

6. Apr 28, 2013

### polpol

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.