Solving a non-elementary integral

Question

$$L=\lim_{\epsilon\to 0^+}\int_{\epsilon}^{2\epsilon}\frac{e^{(x-1)^2}}{x}dx$$ How can one approach this integral? I've tried inserting an extra variable to perform Feynman's trick but it just complicates it more.Kindly give a few hints on how to solve this.

Answer 1

Hint: perform the substitution $x=\epsilon u$ so that $\epsilon$ only appears inside the integrand and not in the bounds. Then you can move the limit inside the integral (be sure to justify this step with your favorite convergence theorem).

Answer 2

Term-by-term integration of the Laurent series... (Conveniently, there's only one disk or annulus of convergence with inner radius $0$ and outer radius $\infty$ and the pole at $x = 0$ is simple.)

\begin{align*} \frac{\mathrm{e}^{(x-1)^2}}{x} &= \frac{\mathrm{e}}{x} - 2 \mathrm{e}+3 \mathrm{e}x - \frac{10 \mathrm{e} x^2}{3} + \frac{19 \mathrm{e} x^3}{6} + \cdots \\ \int_{\epsilon}^{2\epsilon} \; \frac{\mathrm{e}}{x} &{}- 2 \mathrm{e}+3 \mathrm{e}x - \frac{10 \mathrm{e} x^2}{3} + \frac{19 \mathrm{e} x^3}{6} + \cdots \,\mathrm{d}x \\ {} &= \mathrm{e}\ln(2) - 2 \mathrm{e}\epsilon + \frac{9 \mathrm{e} \epsilon^2}{2} - \frac{70 \mathrm{e} \epsilon^3}{9} + \frac{95 \mathrm{e} \epsilon^4}{8} + \cdots \\ &\underset{\epsilon \rightarrow 0^+}{\longrightarrow} \mathrm{e} \ln 2 \text{.} \end{align*}