Proof that this function is bounded How can I prove that $$\int_0^1(1+\epsilon^{-1}e^{-\beta x/p\epsilon})dx\leq C,$$ where $\epsilon$ is a small parameter. In the limit $\epsilon^{-1}$ is unbounded so how is this integral bounded?
 A: Just calculate:
$$\int_0^1 \! (1 + \epsilon^{-1}e^{-\beta x/p\epsilon}) \, dx$$
$$ = 1 + \epsilon^{-1}\int_0^1 \! e^{-\frac{\beta}{p \epsilon}x} \, dx$$
$$ = 1 + \frac{1}{\epsilon}\frac{-p\epsilon}{\beta}(e^{\frac{-\beta}{p\epsilon}} - 1)$$
$$ = 1 + \frac{p}{\beta}(1-e^{\frac{-\beta}{p\epsilon}})$$
This last expression is bounded in absolute value as long as $\epsilon$ does not become arbitrarily negative. Basically the reason the integral stays bounded is that the exponential term becomes small much faster than the $\epsilon^{-1}$ term becomes large as $\epsilon \to 0$ from the right. Notice also that you are only asking for an upper bound
$$\int_0^1 \! (1 + \epsilon^{-1}e^{-\beta x/p\epsilon}) \, dx \le C$$
and such an expression is valid for any $\epsilon$ with $C = 1 + \frac{p}{\beta}$.
A: If I am not mistaken, your integral equals
$$1+\frac{p}{\beta}(1-e^{-\frac{\beta}{p\varepsilon}})$$
Assuming that your parameters are positive, the result follows. 
A: Change variables, $x=\epsilon u$ in your integral:
$$
\begin{align}
\int_0^1\left(1+\epsilon^{-1}e^{-\frac{\beta x}{p\epsilon}}\right)\,\mathrm{d}x
&=\int_0^{1/\epsilon}\left(\epsilon+e^{-\frac{\beta u}{p}}\right)\,\mathrm{d}u\\
&\to1+\int_0^\infty e^{-\frac{\beta u}{p}}\,\mathrm{d}u\\
&=1+\frac{p}{\beta}
\end{align}
$$
as $\epsilon\to0$. So for $\epsilon$ near $0$, the integral is near $1+\frac{p}{\beta}$.
