Divergence of $\int_{0}^{\infty} [ye^{1/(2y)}]^{-1}\text{ d}y$ This came up on a probability exam I took.

Show that  $$\int_{0}^{\infty}\dfrac{1}{ye^{1/(2y)}}\text{ d}y$$ is
  divergent.

In the language of probability, suppose $Y$ has pdf $\dfrac{1}{2y^2}e^{-1/(2y)}$ for $y > 0$. Show that $\mathbb{E}[Y]$ does not exist.
When I saw this question, integration by parts doesn't work too well, nor could I think of an integral for comparison. I still look at this question and am unsure how to deal particularly with the $e^{-1/(2y)}$. 
This does not look like a straight substitution would work either, as we would need a $y^2$ term, rather than just a $y$ term.
Answers should not use anything more than what is taught at a multivariate calculus background (real analysis, outside of inequalities, should be unnecessary).
 A: As $y\to \infty$, $e^{1/2y} \to 1$.  So for some number $N$, $y>N$ assures us that $e^{1/2y} > 1/2.$  Then 
$$\int_1^{\infty} \geq \int_N^{\infty} \geq \int_N^{\infty} \frac{1}{2y} \; dy $$ which diverges.
A: The problem is the integrand when $y\to \infty$. 
Change variable $y=\frac{1}{2 x}$ so $$I=\frac{e^{-\frac{1}{2 y}}}{y}\,dy=-\int\frac{e^{-x}}{x}\,dx$$ When $x$ is small $$\frac{e^{-x}}{x}=\frac{1}{x}-1+\frac{x}{2}+O\left(x^2\right)$$ $$\int\frac{e^{-x}}{x}\,dx\approx \log (x)-x+\frac{x^2}{4}+O\left(x^3\right)$$ 
Edit
In fact, using special functions,  $$I=\int\frac{e^{-\frac{1}{2 y}}}{y}\,dy=-\text{Ei}\left(-\frac{1}{2 y}\right)$$ where appears the exponential integral function. So, $$J=\int_0^a\frac{e^{-\frac{1}{2 y}}}{y}\,dy=\Gamma \left(0,\frac{1}{2 a}\right)$$ where appears the incomplete gamma function. For large values of $a$, the asymptotics is $$J=\log (a)-\gamma +\log (2)+\frac{1}{2
   a}+O\left(\frac{1}{a^2}\right)$$
A: As $y\to\infty$, we have asymptotically
$$
\frac1ye^{-\frac1{2y}}=\frac1y-\frac1{2y^2}+O\left(\frac1{y^3}\right)
$$
Therefore,
$$
\int_1^\infty\frac1ye^{-\frac1{2y}}\,\mathrm{d}y
$$
is divergent since
$$
\int_1^\infty\frac1y\,\mathrm{d}y
$$
is divergent and 
$$
\int_1^\infty\left(-\frac1{2y^2}+O\left(\frac1{y^3}\right)\right)\,\mathrm{d}y
$$
is convergent.
A: First we look at $f(y) = y \cdot e^{\frac{1}{2y}}$. Then $\int_0^\infty f(y)\mathrm{d}y <\infty$ (in the language of probability; we know the expected value of expontential decay probability is finite).
Next, note that $$\int_0^\infty 1 = \int_0 ^\infty \frac{\sqrt{f}}{\sqrt{f}} \leq \left(\int_0^\infty f\right)^\frac{1}{2}\cdot \left(\int_0^\infty f^{-1}\right)^\frac{1}{2}$$ By holder inequality.
Since the left-hand side is $\infty$, the right-hand side must also be $\infty$. Now, we know $\int_0^\infty f(y)\mathrm{d}y <\infty$, and so we conclude $\int_0^\infty f^{-1}(y)\mathrm{d}y =\infty$
