convolution with integration by parts I have a question for an equation from a paper. The paper says,
$$ \frac{\partial c}{\partial t}*g=\int_0^t \frac{\partial c(t-\tau)}{\partial \tau}g(\tau) d\tau=c*\frac{\partial g}{\partial t}+cg_0-c_0g $$
When I use the rule of integration by parts, I got different answer. More specifically, my derivation shows that the right hand side is in negative. Here is my derivation,
$$ \frac{\partial c}{\partial t}*g=c(t-\tau)g(\tau)|_0^t-\int_0^t c(t-\tau)\frac{\partial g(\tau)}{\partial \tau}d\tau=-(c*\frac{\partial g}{\partial t}+cg_0-c_0g ) $$
I have been struggling for this for the whole day. Hope someone can help me. Thanks in advance.
 A: The paper's error is not in the result of the integration but in the integrand. It should be
$$
\frac{\partial c}{\partial t}*g=\int_0^t \frac{\partial c}{\partial t}(t-\tau)g(\tau)\,\mathrm d\tau\;.
$$
Then integrating the first factor with respect to $\tau$ gives you the missing minus sign.
A: Their result can be expressed fairly simply by way of the Laplace transform methods, but given the context of an integration by parts approach I suspect a chain rule implemented to deal with the derivative of c factor (WRT tau) will produce the desired result.
For instance
$\frac{\partial}{\partial \tau} c(t-\tau) = \frac{\partial c(t-\tau)}{\partial (t - \tau)}\frac{\partial (t-\tau)}{\partial \tau} = -\frac{\partial c(t-\tau)}{\partial (t - \tau)}$
Which leads to an IGBP result of:
$\frac{\partial c}{\partial t}*g=-\int_0^t \frac{\partial c(t-\tau)}{\partial (t-\tau)}g(\tau) d\tau=c*\frac{\partial g}{\partial t}+cg_0-c_0g$
Which agrees with the papers outlined result as well as that obtained via Laplace transform methods.
