Show that for f differentiable and bounded by [MATH]\mathbb{R}[/MATH] and [MATH]g[/MATH] integrable
[MATH]\frac{d}{dt}\int_{\mathbb{R}}^{} f(t-r)g(r)dr = \int_{\mathbb{R}} f'(t-r)g(r)dr = \int_{\mathbb{R}}f'(r)g(t-r)dr[/MATH]
and if additionally [MATH]g[/MATH] is differentiable,
[MATH]\frac{d}{dt} \int_{0}^{t}f(t-r)g(r)dr = f(0)g(t)+ \int_{0}^{t}f'(t-r)g(r)dr = g(t)f(0)+ \int_{0}^{t}f(t-r)g'(r)dr [/MATH].
So what I did first was to see that the [MATH]f(t-r)[/MATH] is the only variable dependent on t, so I differentiated this term to get
[MATH]\frac{d}{dt}\int_{\mathbb{R}}^{} f(t-r)g(r)dr = \int_{\mathbb{R}} f'(t-r)g(r)dr[/MATH].
Can I now do the substitution [MATH]u = t-r[/MATH] to get
[MATH]\int_{\mathbb{R}} f'(t-r)g(r)dr = \int_{\mathbb{R}}f'(u)g(t-u)du[/MATH]?
And the second part of the question I am unsure about.
[MATH]\frac{d}{dt}\int_{\mathbb{R}}^{} f(t-r)g(r)dr = \int_{\mathbb{R}} f'(t-r)g(r)dr = \int_{\mathbb{R}}f'(r)g(t-r)dr[/MATH]
and if additionally [MATH]g[/MATH] is differentiable,
[MATH]\frac{d}{dt} \int_{0}^{t}f(t-r)g(r)dr = f(0)g(t)+ \int_{0}^{t}f'(t-r)g(r)dr = g(t)f(0)+ \int_{0}^{t}f(t-r)g'(r)dr [/MATH].
So what I did first was to see that the [MATH]f(t-r)[/MATH] is the only variable dependent on t, so I differentiated this term to get
[MATH]\frac{d}{dt}\int_{\mathbb{R}}^{} f(t-r)g(r)dr = \int_{\mathbb{R}} f'(t-r)g(r)dr[/MATH].
Can I now do the substitution [MATH]u = t-r[/MATH] to get
[MATH]\int_{\mathbb{R}} f'(t-r)g(r)dr = \int_{\mathbb{R}}f'(u)g(t-u)du[/MATH]?
And the second part of the question I am unsure about.