Question: The question is, given $Y\sim N(\mu,\sigma^2)$, how to prove$E[e^{e^Y}]=\infty$?

I tried to look Y as some kind of Ito's process and apply Ito's formula to it but it doesn't make sense. Next I tried to use variable change $u=e^y$ and I still can't prove it. Is there anyway to do that?

Answer: 
Edit: this was posted before the question was edited to reflect the clarification. The following is the answer to a different problem:

If $b<a$, it's the line segment along $x=a$ with length $b$. Otherwise, it's the line segment along $y=b$ with length $a$ (if $a=b$, then they're the same length). You can see this by noting that if you fix the point at (a,b) but move the line, the distance will increase according to the Pythagorean theorem. So the answer is $\min(a,b)$.
Here is my solution.

![enter image description here](https://i.stack.imgur.com/SzHZ0.png)

Your given point is $P(a,b)$, and we let our independent variable be angle $\theta$ as given in the diagram. The other points result from $P$ and $\theta$.

Length $PQ$ is then $a\sec\theta$ and length $PR$ is $b\csc\theta$. The length of segment $QR$ is then

$$L=a\sec\theta+b\csc\theta$$
$$\frac{dL}{d\theta}=a\sec\theta\tan\theta-b\csc\theta\cot\theta$$
$$=a\frac{\sin\theta}{\cos^2\theta}-b\frac{\cos\theta}{\sin^2\theta}$$

Setting this equal to zero and solving we get

$$\frac{\sin^3\theta}{\cos^3\theta}=\frac ba$$
$$\tan\theta=\sqrt[3]{\frac ba}$$

Then

$$QS=a\tan\theta=a\sqrt[3]{\frac ba}$$
$$TR=\frac b{\tan\theta}=b\sqrt[3]{\frac ab}$$

Then using the Pythagorean theorem on triangle $OQR$,

$$L\_{min}=\sqrt{\left(b+a\sqrt[3]{\frac ba}\right)^2+\left(a+b\sqrt[3]{\frac ab}\right)^2}$$
$$=\left(a^{2/3}+b^{2/3}\right)^{3/2}$$

You can fill in the final simplification. This answer works in a quick check using Geogebra.