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?

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 $C = 1 + \frac{p}{\beta}$.
