# Contour integral around essential singularity

1. Aug 10, 2009

### StatusX

Is there a way to perform a contour integral around zero of something like f(z)/z e^(1/z), where f is holomorphic at 0? If you expand you get something like:

$$\frac{1}{z} \left( f(0) + z f'(0) + \frac{1}{2!} z^2 f''(0) + ... \right) \left( 1 + \frac{1}{z} + \frac{1}{2!} \frac{1}{z^2} + ... \right)$$

$$= \frac{1}{z} \left( f(0) + f'(0) + \frac{1}{2!^2} f''(0) + ... \right) + ...$$

So that the result is $2\pi i ( f(0) + f'(0) + \frac{1}{2!^2} f''(0) + ... )[/tex]. Is this correct, and is there a simpler way to write the final answer? If not for the factorials being squared, the term inside the parentheses would just be f(1), but as it is I don't see what I can do with it in general. By the way, the f I'm interested in has complicated poles and branch cuts away from zero, so deforming the contour probably won't help. 2. Aug 10, 2009 ### Count Iblis Yes, that is what you get. A well known example is the Bessel function. If you try to evaluate the integral of exp[i x cos(theta)] d theta from zero to 2 pi, you can rewrite it as the contour integral over the unit circle of: exp[i x/2 (z+1/z)] dz/(i z) So, we've put z = exp(i theta) and then integrating from theta from 0 to 2 pi amounts to integrating z over the unit circle. This is similar to your general form with f(z) = exp(i x/2 z). If you comoute the integral byextracting the residue, you get the series expansion of the Bessel function, which contains factorial squares in the denominators. But what you also can do is find asymptotic expansions for large x, e.g. using the saddle point method. 3. Aug 11, 2009 ### StatusX Ok, well let me give the integral I'm interested in: $$\int_{-1}^1 dx \frac{e^x}{x+b} \tan^{-1} \left( \frac{\sqrt{1-x^2}}{x+a} \right)$$ here a is some arbitrary real number greater than 1, and b is picked so that at x=-b the argument of inverse tan is i (explicitly it's b=(a^2+1)/2a). This looks pretty nasty, but I think it can be done using techniques from complex analysis. In fact, if the e^x isn't there, I've been able to show the answer is: $$\int_{-1}^1 dx \frac{1}{x+b} \tan^{-1} \left( \frac{\sqrt{1-x^2}}{x+a} \right) = \pi \log \left( \frac{a^2}{a^2-1} \right)$$ Basically there's a branch cut from the square root between -1 and 1 and one from the inverse tan running from the pole at x=-b to [itex]-\infty$. The original integral can be written in terms of a contour integral around the first branch cut, and this can be deformed to one around the second. Integrating one way along this branch cut, and subtracting the result going the other way on the other side, we pick up the discontinuity in the function, which is just $\pi/(x+b)$. Its integral diverges, but the divergence is cancelled by contour integral at large R and around the pole at x=-b. I can draw this if anyone's interested.

However, when the e^x is included, the integral along the branch cut becomes that of $\pi e^x/x$ from $-R$ to $-\epsilon$, which I can probably do exactly (at least in the large R and small $\epsilon$ limits), and result at x=b just gets multiplied by $e^{-b}$. But the large R integral seems undoable. Changing variables, it becomes the contour integral around zero of something like:

$$\frac{e^{1/z}}{bz+1} \tan^{-1} \left( \frac{\sqrt{z^2-1}}{az+1} \right)$$

It seems like I need to know all derivatives of the function multiplying e^1/z, and I don't think there's a general expression for them. Does anyone have any ideas on how to perform this final contour integral, or of an alternate way of approaching the original integral?

Last edited: Aug 11, 2009