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

[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?

 

[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?

 

[Office_Shredder]

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)

 

[polpol]

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)

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.

 

[Office_Shredder]

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)

 

[polpol]

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)

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.