- #1
polpol
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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?
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?
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