The theta is comparable to the f in f(x). So to put the integral in similar terms, $\displaystyle \int{e^{au}\theta(u)du}$ is $\displaystyle \int{e^{ax}f(x)dx}$. Can you go from here?
The theta is comparable to the f in f(x). So to put the integral in similar terms, $\displaystyle \int{e^{au}\theta(u)du}$ is $\displaystyle \int{e^{ax}f(x)dx}$. Can you go from here?