I try to calculate the convolution of $\theta(t)\cdot(e^t\theta(1-t))$.
Using the formula
\begin{equation} f*g(t)=\int_{-\infty}^\infty f(t-u)g(u)du \end{equation}
I set $f(t-u)=\theta(t-u)$
and
$g(u)=e^t\theta(1-t)$
So the integral would be:
\begin{equation} \int_{-\infty}^{-1} e^t\theta(1-t)\theta(t-u)du+\int_{-1}^{0}e^t\theta(1-t)\theta(t-u)du+\int_{0}^{\infty}e^t\theta(1-t)\theta(t-u)du \end{equation}
But this gives 1, which is wrong.
Then I tried the formula for casuality:
\begin{equation} f*g(t)=\int_{0}^t f(t-u)g(u)du \end{equation}
and here I got: $-te^t$
which is also wrong.
Where is the error here, in the integral boundaries, or the procedure itself?
Thanks
You're mixing up your variables. The convolution integral should be evaluated as follows:$$\left[e^t\,\theta(1-t)\cdot\theta(t)\right](u)=\int\limits_{-\infty}^{\infty} e^t\,\theta (1-t)\,\theta(u-t)\,dt=e^u+\left(e-e^u\right)\,\theta(u-1)$$