# Convolution product between x² and x³

1. Apr 12, 2014

### Jhenrique

I tryied make the convolution product between x² and x³ by $\int_{- \infty}^{+ \infty} f(u) g(x-u) du$ and the result is an indeterminate form, however, by defintion $\int_{0}^{x} f(u) g(x-u) du$, the result is 1/60 x6. So, $$\int_{- \infty}^{+ \infty} f(u) g(x-u) du \overset{?}{=} \frac{1}{60} x^6$$ ?

2. Apr 12, 2014

### Staff: Mentor

Show us what you did, please.

3. Apr 14, 2014

### Jhenrique

BTW, exist a symmetric algebraic definition for convolution where the graphic of the function f(x) is moved oppositely to g(x) and vice-versa (different of the standard definition, where a function is fixed and the another is moved over the first)? I tryied $\int_{- \infty}^{+ \infty} f(\frac{1}{2}x+u) g(\frac{1}{2}x-u) du$, but this expression isn't equal to the convolution of f and g...

4. Apr 14, 2014

### micromass

OK, so far.

According to you, how does this follow exactly:

5. Apr 14, 2014

### Jhenrique

This equation isn't an affirmation, is a question. In acctually, I'm asking if $\int_{- \infty}^{+\infty} f(u)g(x−u)du$ is equal to $\int_{0}^{x} f(u)g(x−u)du$.

6. Apr 14, 2014

### micromass

I know it's a question. But you're not going to learn much if we solve the question for you. So again, why do you think it's equal? Try to give the proof you're thinking of.

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