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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?

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  • $\begingroup$ It seems to me that you can calculate the integral explicitly, and then see what happens to your $\varepsilon$. $\endgroup$
    – R.T.
    Aug 8, 2013 at 16:39

3 Answers 3

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

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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.

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

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