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# Super simple question about the integral of a convolution watch

1. I want to prove that

So, let's suppose that . Then we can compute:

Now, I guess I have to make a substitution here, but I don't know which one. Could anyone help?
2. u=x-y (to replace x)
3. (Original post by DFranklin)
u=x-y (to replace x)
Don't I have to deal with a ; or can I treat the inner integral separately? I think you are right though; is still , after all.
4. Yes y is still y....
5. (Original post by DFranklin)
Yes y is still y....
Thanks again, Dr. Franklin!

Spoiler:
Show
Although I wonder if you could perhaps post in my thread about the Abel transform...
6. (Original post by RamocitoMorales)
Don't I have to deal with a ; or can I treat the inner integral separately? I think you are right though; is still , after all.
I don't understand your question here, but inside the inner integral, we treat y as constant, since in say R^2, we're integrating over a line of constant y. So z=x-y gives dz=dx
7. (Original post by atsruser)
I don't understand your question here, but inside the inner integral, we treat y as constant, since in say R^2, we're integrating over a line of constant y. So z=x-y gives dz=dx
Yes, Dr. Franklin already said the same. I had already figured that would have sufficed, but sought confirmation. Besides, I didn't include "super simple" in the title to deceive anyone.
8. (Original post by RamocitoMorales)
Yes, Dr. Franklin already said the same.
I didn't understand your question or his response, so I replied anyway.

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