Convolution integral solution

Question

How would I solve the following convolution integral? I would normally be able to solve it if the $t-\tau$ is in the $h$ function, but it is in the exponential instead. $$\int_0^te^{-\frac UB\lambda_L(t-\tau)}h(\tau)d\tau$$

Answer 1

Note that $h(\tau)=1$ for $\tau\in[0,t]$ and the exponential can be broken into two terms, one with $\tau$ and one without $\tau$. So \begin{align} \int_0^t e^{-\frac UB\lambda_L(t-\tau)}h(\tau)\ \mathsf d\tau &= \int_0^t e^{-\frac UB\lambda_Lt}e^{\frac UB\lambda_L\tau}\ \mathsf d\tau\\ &= e^{-\frac UB\lambda_Lt}\int_0^t e^{\frac UB\lambda_L\tau}\ \mathsf d\tau\\ &= \frac B{U\lambda_L}e^{-\frac UB\lambda_Lt}\left(e^{\frac UB\lambda_Lt}-1\right). \end{align}