anuttarasammyak said:
for a>0 ##\beta \neq 0## if not mistaken
I=\frac{e^{-\frac{\alpha^2}{2\beta}}}{a}\ Re \int_0^\frac{\pi}{2} d\theta \ \ e^{-2\beta a^2(cosec\ \theta-i\frac{\alpha}{2\beta a})^2}
For ##\alpha=\beta=0, I=\frac{\pi}{2a}## which is an upper limit of the integral.
[EDIT]
I=\frac{1}{8ai}\int_C \frac{dz}{z} \{ \ \exp[ \frac{B}{(z-z^{-1})^2}+\frac{A}{z-z^{-1}}]+ \exp[ \frac{B}{(z-z^{-1})^2}-\frac{A}{z-z^{-1}}] \ \}
where closed contour C is the unit circle,
A=4a\alpha,\ B=8a^2\beta
but I cannot go further.
With the result of
@renormalize post #6 for A=0
\int_C \frac{dz}{z} \ \exp[ \frac{B}{(z-z^{-1})^2}]=2\pi i \ erfc (\frac{\sqrt{B}}{2})
It seems as if ##erfc (\frac{\sqrt{B}}{2})## is the residue at pole z=0. An interesting relation.
The function ##\exp [\frac{B}{(z-z^{-1})^2}] = \exp [\frac{B z^2}{(1-z^2)^2}]## doesn't seem to be singular at ##z=0##. The poles appear to be
on the contour ##C## at ##z = \pm 1## instead.
Additionally, the result of expanding ##\exp [\frac{B z^2}{(1-z^2)^2}]## and dealing with each term at a time: ##\frac{1}{4ai} \int_C \frac{dz}{z} \exp [\frac{B z^2}{(1-z^2)^2}] = \frac{1}{4ai} \sum_{n=0}^\infty \frac{B^n}{n!} \oint_C \frac{dz}{z} \frac{z^{2n}}{(1-z^2)^{2n}}## couldn't match the Taylor series you would be aiming for, which is:
\begin{align*}
\frac{\pi}{2a} \text{erfc} (\sqrt{B}/2) & = \frac{\pi}{2a} (1-\text{erf} (\sqrt{B}/2) ) = \frac{\pi}{2a} (1- \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{B}/2} e^{-t^2} dt)
\nonumber \\
& = \frac{\pi}{2a} (1-\frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_0^{\sqrt{B}/2} t^{2n} dt)
\nonumber \\
& = \frac{\pi}{2a} (1-\frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^n}{n!} \left[ \frac{t^{2n+1}}{2n+1} \right]_0^{\sqrt{B}/2})
\nonumber \\
& = \frac{\pi}{2a} - \frac{\sqrt{\pi}}{2a} \sqrt{B} \sum_{n=0}^\infty \frac{(-1)^n}{2^n n! (2n+1)} B^n
\end{align*}
because of the factor of ##\sqrt{B}## outside the sum. This means you can't evaluate your complex contour integral by first expanding the exponential. It would be interesting to see how to deal with these issues.
Expanding the exponetial your ##\theta## integral doesn't work as you get divergent integrals:
I agree with your ##\theta## integral. However, putting ##\alpha=0## in your integral representation, and then expanding the exponetial would give:
\begin{align*}
I & = \frac{1}{a} Re \int_0^{\pi/2} e^{-2 \beta a^2 cosec^2 \theta} d \theta
\nonumber \\
& = \frac{1}{a} \sum_{n=0}^\infty \frac{(-1)^n (2 \beta a^2)^n}{n!} \int_0^{\pi/2} cosec^{2n} \theta d \theta
\nonumber \\
& = \frac{\pi}{2a} + \frac{1}{a} \sum_{n=0}^\infty \frac{(-1)^{n+1} (2 \beta a^2)^{n+1}}{(n+1)!} \int_0^{\pi/2} cosec^{2n+2} \theta d \theta
\nonumber \\
& = \frac{\pi}{2a} + 2 a \beta \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(n+1)!} (2 \beta a^2)^n \int_0^{\pi/2} cosec^{2n+2} \theta d \theta
\end{align*}
but the Talyor series expansion for ##\frac{\pi}{2a} \text{erfc} (\sqrt{2 \beta} a)## is ##\frac{\pi}{2a} - \sqrt{2 \pi}\sqrt{\beta} \sum_{n=0}^\infty \frac{(-1)^n}{n! (2n+1)} (2 \beta a^2)^n##. We get the same issue as before, that there is a ##\sqrt{\beta}## outside the sum.
The reason this approach can't work is because all the cosec integrals diverge: First, ##\frac{d}{d \theta} \cot \theta = - cosec^2 \theta## implies
\begin{align*}
\int_0^{\pi/2} cosec^2 \theta d \theta = \left[ - \cot \theta \right]_0^{\pi/2} = \infty
\end{align*}
Then as ##cosec^2 \theta < cosec^{2n+2} \theta## implies that ##\infty = \int_0^{\pi/2} cosec^2 \theta d \theta < \int_0^{\pi/2} cosec^{2n+2} \theta d \theta##.
Noting
\begin{align*}
\infty = \frac{1}{2^{2n}} \int_0^{2\pi} cosec^{2n} \theta d \theta = \oint_C \frac{dz}{z} \frac{z^{2n}}{(1-z^2)^{2n}}
\end{align*}
and ignoring the issue of poles on the contour, the contour integral diverges, which is why expanding the exponential ##\frac{1}{4ai} \int_C \frac{dz}{z} \exp [\frac{B z^2}{(1-z^2)^2}]##doesn't work.
I agree with your ##\theta## integral:
A minor point: I get the same answer as you did for their integral expression: Using ##x=a cosec \theta##
\begin{align*}
I & = e^{-\alpha^2/2\beta} Re \int_a^\infty \frac{e^{-2 \beta (x - i \alpha/2\beta)^2}}{x \sqrt{x^2-a^2}} dx
\nonumber \\
& = e^{-\alpha^2/2\beta} Re \int_{\pi/2}^0 \frac{e^{-2 \beta (a cosec \theta - i \alpha/2\beta)^2}}{a^2 cosec \theta \cot \theta} \cdot - a \cot \theta cosec \theta d \theta
\nonumber \\
& = \frac{e^{-\alpha^2/2\beta}}{a} Re \int_0^{\pi/2} e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta
\end{align*}
I'm not quite getting the same result as you for the next part:
\begin{align*}
I & = \frac{e^{-\alpha^2/2\beta}}{2a} (\int_0^{\pi/2} e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta + \int_0^{\pi/2} e^{-2 \beta a^2 (cosec \theta + i \alpha/2a\beta)^2} d \theta)
\nonumber \\
& = \frac{e^{-\alpha^2/2\beta}}{4a} (\int_0^\pi e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta + \int_\pi^{2\pi} e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta)
\nonumber \\
& = \frac{e^{-\alpha^2/2\beta}}{4a} \int_0^{2\pi} e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta
\nonumber \\
& = \frac{e^{-\alpha^2/2\beta}}{4a i} \oint_C e^{- 2 \beta a^2 (2i(z-z^{-1})^{-1} - i \alpha/2a\beta)^2} \frac{d z}{z}
\nonumber \\
& = \frac{1}{4a i} \oint_C e^{8\beta a^2 [(z-z^{-1})^2 + (z-z^{-1}) \alpha/2a\beta)} \frac{d z}{z}
\nonumber \\
& = \frac{1}{4a i} \oint_C \frac{d z}{z} \exp [8\beta a^2 (z-z^{-1})^{-2} + 4 a \alpha (z-z^{-1})^{-1}]
\nonumber \\
& = \frac{1}{4a i} \oint_C \frac{d z}{z} \exp [\frac{B}{(z-z^{-1})^2} + \frac{A}{(z-z^{-1})}]
\end{align*}
where I have used ##cosec \theta## is symmetric around ##\pi/2##, ##cosec (\theta+\pi)=-cosec \theta##, and put aside subtlies about poles being on the contour.
