# Understanding t bounds on MGF

I'm having trouble understanding what the bounds of the $$t$$ variable are for an mgf. My questions are bolded. Here's an example from a textbook:

Suppose X is a random variable for which the pdf is:

$$f(x) = \begin{cases}e^{-x} & x > 0\\0 & otherwise\end{cases}$$

Determine the mgf of X.

$$\psi(t) = E(e^{tX}) \int_0^\infty e^{tx}e^{-x}dx$$ $$= \int_0^\infty e^{(t-1)x}dx$$

I understand the above, but here is where I am confused. The book then says: "The final integral in this equation will be finite if and only if $$t < 1$$. Therefore, $$\psi(t)$$ is finite only for $$t<1$$."

1) How do you determine that it's only finite for $$t<1$$?

Additionally, it goes on to solve the integral: $$\psi(t) = \frac{1}{1-t}$$, then it states: "Since $$\psi(t)$$ is finite for all values of $$t$$ in an open interval around the point $$t=0$$, all moments of X exist."

2) What does it mean by an open interval around $$t=0$$?

Hint: The integrand $$e^{(t-1)x}$$ has value $$1$$ if $$t=1$$ and is an increasing function of $$x$$ if $$t > 1$$. What do you suppose is the value of the integral when $$t \geq 1$$? In particular, is the value a finite number?