The formula for a convolution is given by
$$f*g(t,x)=\int_{\mathbb{R}}f(t-u)g(u) \ du. \tag1$$
My question is, is the following equally correct?
$$f*g(t,x)=\int_{\mathbb{R}}f(u)g(t-u) du.\tag2$$
I'm asking because in my case I got
$$f(t,x)=\frac{1}{x^2+1} \quad \text{and} \quad g(t,x)=\frac{e^{-x^2/4t}}{\sqrt{4\pi t}}.\tag3$$
So I get two very different integrals.
Yes, since we get: $$ (f*g)(t) = \int_{-\infty}^\infty \ f(t-u)g(u) \ du = -\int_{\infty}^{-\infty} \ f(v)g(v-t) \ dv = (g*f)(t) $$ So the convolution product is commutative.