Moments of $Y=\log X$, for $X$ distributed $U(0,1)$, are not defined? Suppose that $X$ is uniformly distributed on $(0,1)$, and let $Y=\log(X)$. Then it can be shown that the MGF of $Y$ is $M_y(s)=(s+1)^{-1}$ over the domain $R$ exluding $s=-1$.
By expanding $M_Y(s)$, we can show that $E[Y^k]=\dfrac{(-1)^kk!}{s^{2k+1}} $. Since this function is not defined at $s=0$, does this imply that none of the moments of $Y$ are defined?
 A: The MGF of a random variable $Y$ is, in general, $\mathbb E[e^{s Y}]$. Therefore the $k^{\text{th}}$ derivative of $M_Y(s)$ with respect to $s$ should be $\mathbb E[Y^k e^{sY}]$, and evaluating this at $s=0$ gives $\mathbb E[Y^k]$.
There are two ways to continue, which you seem to be confusing. 
First, we could manually take some derivatives of $M_Y(s)$. For example, the first derivative is $-\frac{1}{(1+s)^2}$, which gives $-1$ when we set $s=0$, so $\mathbb E[Y] = -1$. We could keep going, but taking derivatives like this is a pain. If we did anyway, we'd get $(-1)^k \frac{k!}{(1+s)^{k+1}}$ for the $k^{\text{th}}$ derivative.
Second, we can evaluate all of $M_Y(s)$'s derivatives at $0$ by finding the Taylor series. On one hand, we have $$M_Y(s) = 1 - s + s^2 - s^3 + s^4 - \dots + (-1)^k s^k + \dots $$ by the formula for a geometric series. On the other hand, $$M_Y(s) = M_Y(0) + \frac{s^1}{1!} M_Y'(0) + \frac{s^2}{2!} M_Y''(0) + \dots + \frac{s^k}{k!} M_Y^{(k)}(0) + \dots$$ so we must have $$\frac{s^k}{k!} M_Y^{(k)}(0) = (-1)^k s^k$$ and therefore $M_Y^{(k)}(0) = (-1)^k k!$.
