The integral is : $$I(\epsilon)=\int_0^1 \frac{dt}{\epsilon+e^{-1/t}}$$ I need to find the order of $\epsilon$ when $\epsilon\to 0$, such as : $\lim_{\epsilon\to0} \epsilon I(\epsilon)=C$, where C is non-zero.
I have tried the two following ways:
First try: Using the change of variable $t=\frac{1}{x}$, we have: $$\int_0^1 \frac{dt}{\epsilon+e^{-1/t}} =\int_1^{+\infty}\frac{1}{\epsilon+e^{-x}}\cdot \frac{1}{x^2}\, dx\,.$$ Since $0<e^{-x}<1$, for any $x\in(1,+\infty)$, we obtain: $$\frac{1}{\epsilon+1} \int_1^{+\infty} \frac{1}{x^2}\, dx \leq \int_1^{+\infty}\frac{1}{\epsilon+e^{-x}}\cdot \frac{1}{x^2}\, dx \leq \frac{1}{\epsilon} \int_1^{+\infty} \frac{1}{x^2}\, dx .$$ From this result, we only know the upper bound is $O(\frac{1}{\epsilon})$ ; for the lower bound we have no idea.
Second try: By using integration by parts, we get: $$\begin{aligned} \int_0^1 \frac{dt}{\epsilon+e^{-1/t}} &= \int_0^1 \frac{e^{\frac{1}{t}}}{\epsilon e^{\frac{1}{t}} +1}\,dt \\&=-\left[\frac{1}{\epsilon}\ln(\epsilon e^{\frac{1}{t}} +1)t^2\right]_{t=0}^1+ \frac{2}{\epsilon} \int_0^1 t \ln(\epsilon e^{\frac{1}{t}} +1)\,dt\\&= -\frac{1}{\epsilon}\ln(\epsilon e +1) + \frac{2}{\epsilon} \int_0^1 t \ln(\epsilon e^{\frac{1}{t}} +1)\,dt\,. \end{aligned}$$ I have no idea how to continue.
I’m looking forward to your help, thanks!
Write $R = -\log \epsilon$ and note that $R \to \infty$ as $\epsilon \to 0^+$. Then by the substitution $x = Ru$,
$$ I(\epsilon) = \int_{1}^{\infty} \frac{e^{R}}{1 + e^{-(x-R)}} \, \frac{\mathrm{d}x}{x^2} = \frac{e^{R}}{R} \int_{\frac{1}{R}}^{\infty} \frac{1}{1 + e^{-R(u-1)}} \, \frac{\mathrm{d}x}{u^2}. $$
Now it is not hard to check that
$$ \lim_{R\to\infty} \int_{\frac{1}{R}}^{\infty} \frac{1}{1 + e^{-R(u-1)}} \, \frac{\mathrm{d}x}{u^2} = \int_{1}^{\infty} \frac{\mathrm{d}x}{u^2} = 1, $$
and so, we get the following asymptotic equivalence:
$$ I(\epsilon) \sim \frac{e^R}{R} = \frac{1}{\epsilon \log(1/\epsilon)} \quad \text{as} \quad \epsilon \to 0^+.$$
Indeed, the following is the graph of $\epsilon I(\epsilon) \log(1/\epsilon)$ over $(0, 1)$: