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I have to calculate the integral ($b> 0 $)

$$ I(t) = \int_{C_0} d z \exp(- i z t + (i/2b)z^2 ) \prod_{k=1}^N (z - \epsilon_k )^{-i A_k } . $$

Here $\epsilon_k $ are real numbers, and $A_k$ are positive real numbers. The contour $C_0 $ goes from $-e^{i\pi/4} \infty $ to $e^{i\pi/4} \infty $ with all the $\epsilon_k $ on its right hand side. Generally $A_k $ are not ingeters but merely positive. Hence, $\epsilon_k $ are branch points.

In particular, I am interested in the ratio of the modulus of $I(t)$ as $t \rightarrow \pm \infty $. I suspect that the ratio is independent of your choice of the branches, although I am not sure (quite confused by this ugly looking multi-valued function).

enter image description here

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    I suppose it depends on how you define the product $\prod (z-\epsilon_k)^{-iA_k}$, i.e. where you put the branch cuts and what branch you choose. Without the product in the integral you can use the saddle point method to get an asymptotic like $$J(t) := t^{-1} e^{-ibt^2/2} \sqrt{2\pi b},$$ and with the product (defined in a way that makes things easy) I would expect that the asymptotic would be something like $$J(t) \times \exp\{-i(\log b) \sum_k A_k \}.$$ – 2017-01-17
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    I take back the last part of my comment above. The branch cuts of the product seem to become very important, especially if they're in the upper half-plane as the image seems to indicate. – 2017-01-17
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    @AntonioVargas wouldn't it be correct to assume that the integral is dominated from the contributions arising from the branch points? – 2017-01-17
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    @tired, if the branch cuts lie along paths of steepest descent for the exponent then yes, but if not then I think it may change things. – 2017-01-17

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