I have to evaluate the following integral where $N$ is very large and $\lambda_\epsilon=\lambda+i\epsilon$.
\begin{equation}
I=\int_{-\infty}^{\infty} \exp{\left \{ N\left (-m^2 -\frac{1}{2}\ln(\lambda_\epsilon – 2m) \right )\right\}}\mathrm{d}m
\end{equation}
For these kind of integrals I usually have recourse to asymptotic methods, namely the Laplace method. However this methods seems to fail as I don't have a single maximum in my integration range. Moreover the complex number $\lambda_\epsilon$ bugs me a bit.
One told me to evaluate this integral using hypergeometric functions using an infinitesimal $\epsilon$. However I am completely unfamiliar with this.
Any advice or remarks are highly appreciated! Thank you.
Best Answer
I assume that $\ln$ denotes the principal value of the logarithm. $I$ can be evaluated in closed form in terms of two ${_1 \hspace {-1px} F_1}$ functions, but getting the asymptotics for those is not easy. Instead, we can apply the steepest descent method to the integral directly.
Assume $\epsilon > 0$ and $\lambda > 0$. The stationary points of $\phi(m) = -m^2 - \ln(\lambda + i \epsilon - 2 m)/2$ are at $$m_{1, 2} = \frac {\lambda + i \epsilon \pm \sqrt {(\lambda + i \epsilon)^2 - 4}} 4.$$ The level curves of $\operatorname {Re} \phi$ look like this (the values increase from blue to magenta to red):
Because of the branch cut, we cannot deform the contour $(-\infty, \infty)$ to go through $m_1$ while avoiding the red regions. Therefore, we need to take a contour that goes through $m_2$ in the direction from left to right. This corresponds to $m = m_2 + \xi \sqrt {-2/\phi''(m_2)}$ near the saddle point $m_2$. Then $$I \sim \sqrt {-\frac 2 {\phi''(m_2)}} \int_{\mathbb R} e^{N (\phi(m_2) - \xi^2)} d\xi = \sqrt {-\frac {2 \pi} {N \phi''(m_2)}} \, e^{N \phi(m_2)}, \quad N \to \infty.$$