5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What did you try? $\endgroup$ – Jonathan H Aug 6 '13 at 7:14
  • $\begingroup$ 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... $\endgroup$ – mmal Aug 6 '13 at 8:28
  • $\begingroup$ Answer: $-e^{-1}\pi i$. (limit $\epsilon \to 0^+$, the negative on the other side) $\endgroup$ – GEdgar Aug 6 '13 at 12:46
  • $\begingroup$ @mmal Why not try $y\to x+1$ instead? $\endgroup$ – Jonathan H Aug 6 '13 at 18:12
7
$\begingroup$

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,

enter image description here

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.