# Convulation and integration

1. Jan 2, 2014

### Jhenrique

I was verifying that:

http://en.wikipedia.org/wiki/Convolution#Integration

But, I'd like to know if this equality is true:
$$\int f(x)g(x)dx \overset{?}{=} \int f(x)dx \ast \int g(x)dx$$

Subquestion: this theorem above can be applied in summation too?
$$\sum_{x}\left (f(x)\ast f(x) \right )\overset{?}{=}\left(\sum_{x} f(x)\right)\left (\sum_{x} g(x) \right )$$
$$\sum_{x}f(x)g(x) \overset{?}{=} \sum_{x}f(x) \ast \sum_{x}g(x)$$
Thanks!

2. Jan 2, 2014

### pwsnafu

The easy way to check is differentiate both sides. What do you know about the derivative of the convolution.

3. Jan 2, 2014

### R136a1

But $\int f(x) dx$ and $\int g(x)dx$ are just real numbers, no? What is the convolution of two numbers?

4. Jan 2, 2014

### Jhenrique

That:

5. Feb 25, 2014

### Jhenrique

No! The result is a function!

Results: f(x) g(x) = f(x) * ∫ g(x) dx

But, my ask remains! My as ask now is f(x) g(x) is equal to f(x) * ∫ g(x) dx ???

Last edited: Feb 25, 2014