Integral of a complex gaussian + inverse gaussian integral Why does this identity holds ?
$$
I(a, b) := \int_0^\infty dx e^{-ia^2x^2 -i b^2/x^2} = \frac{1}{2|a|} \sqrt{\frac{\pi}{i}} e^{-2i |a| |b|}
$$
It is given in the book Quantum Field theory by T. Padmanabhan, eq. 1.82 page 22.
I am not sure how to tackle this integral. I know that the prefactor can be found by looking at the value of $I(a, b=0)$. Furthermore, looking at the function $J(a, b) = |a| I(a, b)$, it can be found to only depend on $|ab|$, that is $J(a, b) = \int_0^\infty dx \exp(-ix^2 -i|ab|^2/x^2)$.
 A: Use the fact that the function is even to get
$$\int_{0}^\infty e^{-ia^2x^2-i\frac{b^2}{x^2}}\:dx = \frac{1}{2}\int_{-\infty}^\infty e^{-ia^2x^2-i\frac{b^2}{x^2}}\:dx$$
Then use the Glasser Master Theorem
$$\int_{-\infty}^\infty f\left(x-\frac{k}{x}\right)dx = \int_{-\infty}^\infty f(x)dx$$
for $k>0$ after completing the square. Can you take it from here?
A: So to start, we need to clarify that your result is only valid in the case where $a$ and $b $ are real numbers, and not complex.  It is possible to work this out that this is out from the context, but it took me a few minutes.
 I'm going to make the further restriction that $a$ and $b$ are positive. I think it should be relatively straightforward to reinstate the absolute value signs to extend to $a$ and $b$ on the whole real line if you need to do that afterwards.
Then let's consider
 $$
I(a, b) := \int_0^\infty dx e^{-ia^2x^2 -i b^2/x^2}
$$
As you suggest, first look at the case where $b = 0$. Then this is a standard Gaussian integral with a purely imaginary argument, and we can get the result from an integral table, or using contour integration, that
$$
I(a, 0) = \frac{1}{2a} \sqrt{\frac{\pi}{i}}
$$.
Now we'll proceed to differentiate under the integral sign with respect to $b$, and construct a differential equation for $I$ which we can solve. So:
 $$
\partial_b I(a, b) = -2ib \int_0^\infty \frac{dx}{x^2} e^{-ia^2x^2 -i b^2/x^2}
$$
Change variables to $y = \frac{b}{a x}$, noting that this gives $dy = -\frac{b}{ax^2}dx$, and interchanges the $0$ and the $\infty$ in the limits. Then we see that
 $$
\partial_b I(a, b) = 2ia \int_\infty^0 dy e^{-ib^2/y^2 -i b^2 y^2}
$$
Swapping the limits of the integral and introducing a minus sign, we recognise that this is just the original function, $I(a,b)$. Hence we see that $I$ satisfies the equation
 $$
\partial_b I(a, b) = -2ia I(a,b)
$$
We also calculated the initial condition when $b=0$ already, so we can integrate this equation using that $f'(x) = c f(x), f(0) =A$ gives $f(x) = A e^{c x}$, and  we arrive at
 $$
I(a, b) := \int_0^\infty dx e^{-ia^2x^2 -i b^2/x^2} = \frac{1}{2a} \sqrt{\frac{\pi}{i}} e^{-2i a b}
$$
as required.
