Let $ \alpha \le 0 $ and $\sigma > 0$ .

I want to choose a contour, including $ [\sigma - iR, \sigma+iR] $ , such that i can apply Cauchy's Residue theorem and evaluate:

$$ \lim_{R \rightarrow \infty} \int\limits_{\sigma - iR}^{ \sigma+iR } \frac{\exp(\alpha z)}{(z^2 + 1)} dz$$

by contour integration. The case for $\alpha > 0 $ is nice i think, as you can use a semi-circle, but i cant seem to find a suitable contour for this case.

EDIT: to be specific when we use a semi-circle then the integral over the circular path tends to zero in the limit for $\alpha > 0 $ , but this does not seem to happen in the above case.

  • $\begingroup$ How do you expect to apply CRT to a complex integral on a straight line in the complex plane?? We usually do this to calculate real integrals by means of a closed contour in the complex plane... $\endgroup$ – DonAntonio Dec 28 '13 at 15:36
  • $\begingroup$ see my edit, in the limit the only non-zero section of our contour integral is that integrated over the line. it worked for the earlier case, unless i am completely confusesd. $\endgroup$ – KJC Dec 28 '13 at 15:38

You can choose a semicircle in the half-plane $\operatorname{Re} z \geqslant \sigma$. Then $e^{\alpha z}$ is bounded on the contour - $\operatorname{Re} (\alpha z) = \alpha\operatorname{Re} z \leqslant \alpha\sigma \leqslant 0$, and $\lvert e^{\alpha z}\rvert = e^{\operatorname{Re} (\alpha z)}$ - and the integral over the semicircle tends to $0$ for $R \to \infty$. Since the contour encloses no singularity, the contour integral is $0$, and hence

$$\int_{\sigma - i\infty}^{\sigma+i\infty} \frac{e^{\alpha z}}{z^2+1}\,dz = 0$$

for $\sigma > 0$ and $\alpha \leqslant 0$.

It is worth noting that

$$I(\alpha) = \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{e^{\alpha z}}{z^2+1}\,dz = \begin{cases}\quad 0 &, \alpha \leqslant 0\\ 2\pi i\sin \alpha &, \alpha \geqslant 0 \end{cases}$$

does not depend smoothly on $\alpha$, although the integrand does. That should, however, not be too surprising since the differentiated integrand $\dfrac{ze^{\alpha z}}{z^2+1}$ is not integrable anymore, and the non-differentiability in $0$ manifests itself by the need to switch the half-plane in which the contour is closed to have the integral over the auxiliary part tend to $0$.

  • $\begingroup$ could you show that $e^{ \alpha z} $ is bounded on this contour? im not seeing that. $\endgroup$ – KJC Dec 28 '13 at 15:50
  • $\begingroup$ Added the estimate for that. $\endgroup$ – Daniel Fischer Dec 28 '13 at 15:58
  • $\begingroup$ excellent explanation $\endgroup$ – KJC Dec 28 '13 at 22:52

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