# Computing the integral $\lim\limits_{\epsilon\to 0} \int_{-2}^{0} \frac{e^{1/x(x+2)}}{x+1+i\epsilon}$

I got stuck when calculating of this expression

$$\lim_{\epsilon\rightarrow 0} \int_{-2}^{0} \frac{e^{\frac{1}{x(x+2)}}}{x+1+i\epsilon}$$

I will be grateful for the advice.

• What did you try? – Jonathan H Aug 6 '13 at 7:14
• After removing $i\epsilon$ term from denominator and change of variables $y\rightarrow \frac{1}{x(x+2)}$ we get two (unpleasant I think) indefinite integrals... – mmal Aug 6 '13 at 8:28
• Answer: $-e^{-1}\pi i$. (limit $\epsilon \to 0^+$, the negative on the other side) – GEdgar Aug 6 '13 at 12:46
• @mmal Why not try $y\to x+1$ instead? – Jonathan H Aug 6 '13 at 18:12

Let $I(\epsilon)$ denote the integral inside the limit. Using the change of variable $z = x+1+i\epsilon$, we have

$$I(\epsilon) = \int_{-1+i\epsilon}^{1+i\epsilon} \exp\left(-\frac{1}{1-(z-i\epsilon)^2}\right) \, \frac{dz}{z}.$$

Now let

$$f_{\epsilon}(z) = \frac{1}{z} \exp\left(-\frac{1}{1-(z-i\epsilon)^2}\right).$$

and assume that $\epsilon > 0$. Then by considering a rectangular contour consisting of vertices $\pm 1$ and $\pm 1 + i\epsilon$ with an upper semicircular indent $C^{-}_{\eta}$ of radius $\eta > 0$ at the origin,

we can write

$$I(\epsilon) = \int_{-1+i\epsilon}^{-1} f_{\epsilon}(z) \, dz + \int_{-1}^{-\eta} f_{\epsilon}(z) \, dz + \int_{C^{-}_{\eta}} f_{\epsilon}(z) \, dz + \int_{\eta}^{1} f_{\epsilon}(z) \, dz + \int_{1}^{1+i\epsilon} f_{\epsilon}(z) \, dz.$$

Now it is easy to confirm that $\left| f_{\epsilon}(z) \right| \leq 1$ for $\Re z = \pm 1$. Thus by taking $\epsilon \to 0^{+}$, it follows that

$$\lim_{\epsilon \to 0^{+}} I(\epsilon) = \int_{-1}^{-\eta} f_{0}(z) \, dz + \int_{C^{-}_{\eta}} f_{0}(z) \, dz + \int_{\eta}^{1} f_{0}(z) \, dz.$$

But since $f_{0}(z)$ is an odd function, the first integral and the last integral cancel out, yielding

$$\lim_{\epsilon \to 0^{+}} I(\epsilon) = \int_{C^{-}_{\eta}} f_{0}(z) \, dz.$$

Now taking $\eta \to 0$, we have

$$\lim_{\epsilon \to 0^{+}} I(\epsilon) = -i\pi \, \mathrm{Res}_{z=0} \, f_{0}(z) = -\frac{i\pi}{e}.$$

Similar consideration yields

$$\lim_{\epsilon \to 0^{-}} I(\epsilon) = \frac{i\pi}{e}.$$