Question: Provided $\xi \ll 1$, find the first three terms of the asymptotic expansion of the integral $$I(\xi) = \int_{\xi}^{\infty}\frac{e^{-\alpha t}}{t}dt.$$
My approach: Assuming $\alpha>0$ and substituting $x = \alpha t$ in $I(\xi)$, we have $$I(\xi) = \int_{\alpha \xi}^{\infty}\frac{e^{-x}}{x}dx.$$ Now, since $\xi \ll 1$ and $\alpha$ is fixed, implies $\epsilon = \alpha \xi \ll 1$. Therefore, we are left to find the asymptotic expansion of $$J(\epsilon) = \int_{\epsilon}^{\infty}\frac{e^{-x}}{x}dx$$ as $\epsilon \to 0^+$. Differentiating both sides of the above equation with respect to $\epsilon$ and noting that $\epsilon \to 0^+$, we have $$\frac{dJ}{d\epsilon} = -\frac{e^{-\epsilon}}{\epsilon}\sim -\frac{1}{\epsilon}\left[1-\epsilon+O(\epsilon^2)\right]=-\frac{1}{\epsilon}+1+O(\epsilon).$$ This implies that $$J(\epsilon) \sim C - \ln(\epsilon) + \epsilon + O(\epsilon^2),$$ which in turn implies that $$\lim_{\epsilon\to 0^+}\left[J(\epsilon) + \ln(\epsilon)\right] = \lim_{\epsilon\to 0^+} \left(C + \epsilon -\frac{1}{4}\epsilon^2 +\frac{1}{18}\epsilon^3 - \cdots \right) = C.$$ Therefore, $$C = \lim_{\epsilon\to 0^+}\left(\int_{\epsilon}^{\infty}\frac{e^{-x}}{x} + \ln(\epsilon)\right) = -\gamma,$$ where $\gamma$ is Euler's constant. Therefore, $I(\xi) \sim -\ln(\xi) + [-\gamma - \ln(\alpha)] + \alpha\xi.$
Can someone please let me know if this is the correct way to solve this problem. Also, please let me know if there's any other way to solve this problem. Thanks in advance!
It is known that $\gamma$ has the following integral representation
$$ \gamma=\int_0^1{1-e^{-x}\over x}\mathrm dx-\int_1^{+\infty}{e^{-x}\over x}\mathrm dx, $$
so we have
\begin{aligned} J(\epsilon) &=\int_\epsilon^1{e^{-x}-1+1\over x}\mathrm dx+\int_1^{+\infty}{e^{-x}\over x}\mathrm dx \\ &=\int_\epsilon^1{e^{-x}-1\over x}\mathrm dx+\ln\frac1\epsilon+\int_1^{+\infty}{e^{-x}\over x}\mathrm dx \\ &=\ln\frac1\epsilon-\gamma-\int_0^\epsilon{e^{-x}-1\over x}\mathrm dx \end{aligned}
For the remaining integral, we can subsitute in the power series
$$ {e^{-x}-1\over-x}=\sum_{n\ge1}{(-x)^{n-1}\over n!} $$
and deduce that
$$ J(\epsilon)=\ln\frac1\epsilon-\gamma-\sum_{n\ge1}{(-\epsilon)^n\over n\cdot n!}. $$