How Do You Solve Complex Contour Integrals with Trigonometric Substitutions?

[demoncore]

I am attempting to calculate the following integral.

$$\frac{1}{2\pi i}\int_C \frac{du}{u^2} exp({-\frac{(q - \frac{q_0}{2i} (u - u^{-1}))^2}{2\sigma^2}})$$

Taken over the unit disk. I first make the substitution $$z = q - \frac{q_0}{2i} (u - u^{-1})$$ So,
$$dz = -\frac{q_0}{2i}(1 + u^{-2})du$$

When I attempt to back-substitute in for u, however, I find the following expression:

$$u = \frac{(q - z)i \pm \sqrt{q_0^2 - (q - z)^2}}{q_0}$$
I am not sure where to proceed from here. Arbitrarily choosing one or the other solution for u doesn't seem to give me the correct answer. Any help would be appreciated.

 

[ecastro]

Have you tried deconstructing your ##e## to multiple terms? I think you could take a term outside the integral.

 

[Ray Vickson]

I am attempting to calculate the following integral.

$$\frac{1}{2\pi i}\int_C \frac{du}{u^2} exp({-\frac{(q - \frac{q_0}{2i} (u - u^{-1}))^2}{2\sigma^2}})$$

Taken over the unit disk.

If you set ##u = e^{i \theta}## you can write the integral (call it ##J\,## ) in the form
[tex] J = \frac{1}{2 \pi i} \int_0^{2 \pi} i e^{i \theta} e^{-2 i \theta} E(\theta) \, d \theta,\\
E(\theta) = \exp \left(- \frac{1}{2 \sigma^2} \left(q - q_0 \sin(\theta) \right)^2 \right)[/tex]
Thus, ##J = J_r+ i J_i##, where
[tex] J_r = \frac{1}{2 \pi} \int_0^{2 \pi} \cos(\theta) E(\theta) \, d \theta, \\
J_i = -\frac{1}{2 \pi} \int_0^{2 \pi} \sin(\theta) E(\theta) \, d \theta[/tex]
Note that ##J_r = 0## because the integrand is asymmetric about ##\theta = \pi/2## on the interval ##[0,\pi]## and is asymmetric about ##\theta = 3 \pi/2## on the interval ##[\pi, 2\pi]##. However, ##J_i \neq 0##. In numerous numerical examples we find ##J_i < 0##.

I doubt that ##J_i## has a simple (if any) elementary formula, but numerical integration works well on it, especially if one computes the parts for ##[0,\pi]## and ##[\pi, 2\pi]## separately. Alternatively, this separation can be done analytically, to get
[tex] J_i = \frac{1}{2 \pi} \int_0^{\pi} \sin(\theta) F(\theta), \, d \theta, \\
F(\theta) = \exp \left(- \frac{1}{2 \sigma^2} (q + q_0 \sin(\theta))^2 \right) -
\exp \left(- \frac{1}{2 \sigma^2} (q - q_0 \sin(\theta))^2 \right) [/tex]