Expectation of first-passage-time of a diffusion process with negative drift Take  the stochastic process $X_0 = 0$ and $X_t = \nu t + \sigma W_t$  where $W_t$ is standard Brownian motion and $\nu$ is a drift which may be negative (this is the key complexity to the question)
Let $\alpha > 0$ be a fixed level, then define the first passage time as the random variable: $T = \inf\{ 0 < t \mid X_t=\alpha \}$.
Then, given some $r > 0$, what is the following expectation?
$$
U \equiv \mathbb{E}\left(e^{-r T} \right)
$$
A related question may be: what is the moment generating function for the $T$ stopping time?  Note that I do not care about the pdf or cdf of $T$, just the above expectation.

Solving the $\nu \geq 0$ case:
I believe this turns out to be easy as this follows the example in https://en.wikipedia.org/wiki/Inverse_Gaussian_distribution#Relationship_with_Brownian_motion .  Hence, $T$ is distributed as an Inverse Gaussian: $T\sim IG(\tfrac\alpha\nu, \tfrac {\alpha^2} {\sigma^2})$ in that notation.
Moreover, the moment generating function for the $IG(\mu,\lambda)$ is $$M(t;\mu,\lambda) \equiv \exp\left[{\frac{\lambda}{\mu}\left(1-\sqrt{1-\frac{2\mu^2t}{\lambda}}\right)}\right]$$
Finally, note that the expectation I gave is exactly the definition of the MGF at $t=-r$, so we have our answer,
$$
\mathbb{E}\left(e^{-r T} \right) = M\left(-r;\tfrac\alpha\nu, \tfrac {\alpha^2} {\sigma^2}\right)
$$

Solving the $\nu < 0$ case:  If you look carefully at theorems such as in Karlin  and Taylor "A First Course in Stochastic Processes" page 362 and theorem 5.3, or the wikipedia page above, they always assume that $\nu \geq 0$.  As Karlin and Taylor puts it: "When $\nu < 0$, $T$ has a defective probability distribution, that is $T$ is infinite with positive probability".
However, to find $\mathbb{E}\left(e^{-r T}\right)$ we don't really need to find the probability distribution.  Is the moment generating function (and hence the direct solution for $U$) still defined?  Is the formula the same as above?
 A: The answer is
$$ E[e^{-rT}] = exp\left\{\frac{\mu+\sqrt{\mu^2+2\sigma^2r}}{\sigma^2} \alpha\right \} $$ if $\alpha<0$, and
$$ E[e^{-rT}] = exp\left\{\frac{\mu-\sqrt{\mu^2+2\sigma^2r}}{\sigma^2} \alpha \right\} $$ if $\alpha>0$
To establish these formulas one can follow the approach from Karlin and Taylor. Since Karlin and Taylor provide the derivations for the case $\alpha>0$, I derive the formula for the case $\alpha<0$.
Let $V_{\lambda}(t)=exp\{\lambda X(t) - q(\lambda)t \}$ where $q(\lambda) = \mu \lambda+\frac{1}{2}\sigma^2\lambda^2$. Note $V_{\lambda}(t)$ is a martingale. 
Consider $V_{\lambda}(t \wedge T )$. Then $V_{\lambda}(t \wedge T )$ is also a martingale and is bounded for $\lambda<0$. We assume then that $\lambda<0$, and from the above observations we have,
$$ E[V_{\lambda}(t \wedge T )] = E[V(0)] = 1  $$ 
Moreover
$$\lim_{t \rightarrow \infty} V_{\lambda}(t \wedge T) =  \left\{ \begin{array}{ll} 
exp\{\lambda T - q(\lambda) T & \text{  if  } T<\infty \\
0  & \text{ if } T=\infty
\end{array}
 \right.  $$
Therefore,
$$\lim_{t \rightarrow \infty} E[V_{\lambda}(t \wedge T )]= E\left[\lim_{t \rightarrow \infty} V_{\lambda}(t \wedge T )\right] = e^{\lambda \alpha}E[e^{-q(\lambda)T}] $$
From the above derivations it follows that
$$1=e^{\lambda \alpha}E[e^{-q(\lambda)T}]$$ or
$$e^{-\lambda \alpha}=E[e^{-q(\lambda)T}]$$
Set $q(\lambda)=r$ and solve the resulting quadratic to obtain
$$ \lambda=\frac{-\mu \pm \sqrt{\mu^2+2\sigma^2 r}}{\sigma^2}$$ 
Since we require $\lambda<0$ the correct root is the negative one. Thus, we obtain
$$ E[e^{-rT}]=\exp \left\{ \frac{\mu + \sqrt{\mu^2+2\sigma^2 r}}{\sigma^2} \alpha \right\}$$ 
Finally, note that if $\mu>0$ then $\lim_{r \rightarrow \infty} E[e^{-rT}] \neq 1$ which implies that the associated distribution function is defective, while if $\mu>0$ then $\lim_{r \rightarrow \infty} E[e^{-rT}]=1$ and the associated distribution is not defective (Theorem 2 in Feller that  Comic Book Guy referred to).
