I'm trying to evaluate the Laplace transform for the Uniform distribution; i.e., $U(0,\tau)$.
Let's denote the Laplace transform of the distribution as $L(s)$, so:
$$ L(s) = \frac{1 - e^{-s\tau}}{s} $$
To get the MGF from the Laplace transform, we need to replace $ s $ with $ -s $:
$$ M(t) = L(-t) = \frac{1 - e^{t\tau}}{t} $$
Now, to find the expected value (mean) of the distribution, we calculate the first derivative of the MGF at $t=0$:
$$ E[X] = M'(0) $$
Taking the derivative of $ M(t) $ gives:
$$ M'(t) = \tau e^{t\tau} - \frac{1 - e^{t\tau}}{t^2} $$
At this point, we have an indeterminate form when $t=0$, so we must apply L'Hopital's rule and take the second derivative:
$$ M''(t) = \tau^2 e^{t\tau} - \frac{2\tau e^{t\tau}}{t} + \frac{2(1 - e^{t\tau})}{t^3} $$
Evaluating the second derivative at $t = 0$ results in:
$$ M''(0) = \tau^2 - \frac{2\tau}{0} + \frac{2(1 - 1)}{0^3} $$
Again, this computation is undefined.
How can you compute this?
First you need to fix you expression of $M(t)$. Indeed,
$$M(t) = \mathbb E\left[e^{tX}\right] = \frac{e^{\tau t} - 1}{t}$$
then,
$$M'(t) = \frac{\tau e^{\tau t}}{t}- \frac{e^{\tau t} - 1}{t^2} = \frac{1 - e^{\tau t}(1 - \tau t)}{t^2} = \frac{f(t)}{g(t)}$$
Now to compute the limit as $t\to 0$ you need to differentiate the numerator and the denominator:
$$\frac{f'(t)}{g'(t)} = \frac{-\tau e^{\tau t}\left(1-\tau t\right) + e^{\tau t}\tau}{2t} = \frac{\tau te^{\tau t}}{2t} = \frac{\tau}{2}e^{\tau t}\to \frac{\tau}{2}$$