A problem on my textbook on complex analysis requires me to calculate the line integral $$ I := \int_C \exp({\alpha z})\frac{dz}{z^2}, $$ where $C=\{a+iy\mid -\infty < y < \infty\}$, $a \in \mathbb R$ and $\Re \alpha \neq 0$.

I have two problems regarding this.

  1. My textbooks defines line integrals only for curves that is an image of some function from a real closed interval into $\mathbb C$. I suppose $C$ is not of this form, and the direction of $C$ is unspecified. How is this integral defined usually?

  2. I decided that this formula means $$ \int_{-\infty}^{\infty} \exp(\alpha(a+iy))\frac{idy}{(a+iy)^2}.$$

I tried to calculate this integration by real variable by residue calculus. I replaced $\alpha y$ with $x$ and got $$ I =i\exp(\alpha a)\alpha\int\frac{e^{ix}}{(\alpha a + ix)^2}dx, $$ and then by a well-known technique $$ I = i\exp(\alpha a)\alpha\cdot 2\pi i \mathrm{Res}\left(ia\alpha, \frac{e^{ix}}{(\alpha a + ix)^2}\right). $$ According to my calculation the residue is equal to $$ \frac{1}{(2-1)!}\frac{d}{dz}\big|_{z=ia\alpha}e^{iz} = ie^{-a\alpha} $$ and finally $$ I = -2\pi \alpha i. $$

However, the solution section of the textbook says that $I=0$. Where did I went wrong?

EDIT: By a well-known technique I mean a method that can be applied to integrals of the form $$ J:=\int \frac{P(x)e^{ix}dx}{Q(x)}, $$ where $P$ and $Q$ are polynomials such that $\deg P - \deg Q \le -2$. (I believe this appears somewhere in the Ahlfors' book.) One can take an integration path to be a large half circle in the upper half plane plus a curve from left to right on the real axis, except around the poles, where the curve goes down in the shape of lower half circle. One can show the integration along the large upper half circle tends to zero as its radius becomes large, and the integration along the small lower half circles tends to $\pi i$ times the residue around the poles as you make their radii small. In conclusion, we have $$ J = 2\pi i \sum_{\Im a > 0} \mathrm{Res}(a, Pe^{ix}/Q) + \pi i \sum_{\Im a = 0} \mathrm{Res}(a, Pe^{ix}/Q). $$

EDIT 2: I noticed that the method above does not work when $P(z)/Q(z)$ has poles of order 2 or more on the real axis. Thus Question 2 is solved; I am looking forward to an answer to Question 1, though.

  • 1
    $\begingroup$ If you're applying a "well-known technique," what is the closed contour on which you're applying the Residue Theorem, and have you checked that the rest of the integral goes to $0$ and what poles are inside the contour? $\endgroup$ – Ted Shifrin Jun 17 '13 at 15:40
  • $\begingroup$ I guess the problem lies on the way you apply the "well-known technique", you have to write it out for us to check it for you. As Ted said, my guess is that you didn't even make sure that the rest of the line integral goes to $0$ when you apply contour integration and always assume it gives the residue of poles enclosed. $\endgroup$ – Secret Math Jun 17 '13 at 16:30

Your interpretation of the line integral is correct; one way to define it is as the limit as $M$ and $N$ go to $\infty$ of the integral along the vertical line segment $C_{M,N}$ from $a - iM$ to $a + iN$ (and this latter integral is covered by your text's definition).

As you surmised, this leads to the integral $$\int_{-\infty}^{\infty} e^{\alpha(a + i y)} \dfrac{i dy}{(a + iy)^2}.$$

  • $\begingroup$ Thank you. Now I know that the original way I calculated the integral was incorrect, but don't know how to do it correctly. Could you help me in this regard? $\endgroup$ – Pteromys Jun 21 '13 at 11:01

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.