Integral $\int_0^\infty \exp(ia/x^2+ibx^2)dx$ Compute the integral:
\begin{equation}
\int_0^\infty \exp\left(\frac{ia}{x^2}+ibx^2\right)\,dx
\end{equation}
for $a$, $b$ real and positive. I tried complex variables, but don't really know how to handle this.
 A: Let's try to obtain the Laplace integral (from a proof from Albano, Amdeberhan, Beyerstedt and Moll see $5.$) :
$$\tag{1}\int_0^\infty e^{-\frac a{x^2}-bx^2}dx=\sqrt{\frac{\pi}{4b}}e^{-2\sqrt{ab}}$$
Let's suppose $a:=u^2$ and $b:=v^2$ and compute :
$$\tag{2}I=\int_0^\infty e^{-\left(\frac ux-vx\right)^2}dx=e^{2\sqrt{ab}}\int_0^\infty e^{-\frac a{x^2}-bx^2}dx$$
The substitution $\,x:=\dfrac u{v\,z}\,$ gives (after reordering) :
$$I=\frac uv\int_0^\infty e^{-\left(\frac uz-vz\right)^2}\frac{dz}{z^2}$$
adding these equations we get :
$$2\,I=\frac 1v\int_0^\infty e^{-\left(\frac uz-vz\right)^2}\left(\frac u{z^2}+v\right)dz$$
Set $\ \displaystyle t:=\frac uz-v\,z\ $ to get :
$$I=\frac 1{2v}\int_{-\infty}^\infty e^{-t^2}dt=\frac{\sqrt{\pi}}{2v}$$
That we may substitute in $(2)$ to get $(1)$ as wished.
From analytic continuation you may obtain the integral you wanted (this is an argument for physicists! ;-)).
You may see this problem too from the point of view of Dawson integral of the complementary error function $\operatorname{erfc}$ (see $7.7.7$ and $7.7.8$ for the specific case $x=0$).
Since you are interested by the imaginary case this paper of Styer may interest you (he considers the error function as previously but with imaginary terms for the problem $6-13$ from the book). 
This detailed errata page from Styer for Feynman Hibb's book is related too.
A: The integral is the same as 
$$e^{2i \sqrt{ab}}\int_0^{\infty} e^{i ({\sqrt{a} \over x} - \sqrt{b}x)^2}\,dx$$
Letting $x = \sqrt[4]{a \over b}$ $y$, this integral becomes
$$ \sqrt[4]{a \over b} e^{2i \sqrt{ab}}\int_0^{\infty}e^{i\sqrt{ab}(y - {1 \over y})^2}\,dy$$
Changing variables from $y$ to ${1 \over y}$ in the integral gives
$$\int_0^{\infty}e^{i\sqrt{ab}(y - {1 \over y})^2}\,dy = \int_0^{\infty}e^{i\sqrt{ab}(y - {1 \over y})^2}{1 \over y^2}\,dy$$
So the left hand side of the last equation is equal to the average of the left and right hand sides. In other words, 
$$\int_0^{\infty}e^{i\sqrt{ab}(y - {1 \over y})^2}\,dy = {1 \over 2} \int_0^{\infty}e^{i\sqrt{ab}(y - {1 \over y})^2}(1 + {1 \over y^2})\,dy $$
Changing variables here to $z = {y - {1 \over y}}$ gives
$$\int_0^{\infty}e^{i\sqrt{ab}(y - {1 \over y})^2}\,dy = {1 \over 2} \int_{-\infty}^{\infty} e^{i\sqrt{ab}\,z^2}\,dz$$
Letting $z = {1 \over \sqrt[4]{ab}}u$ this is the same as
$${1 \over 2\sqrt[4]{ab}} \int_{-\infty}^{\infty} e^{iz^2}\,dz$$
This is the famous Fresnel integral, and the above has value ${\displaystyle{e^{i\pi \over 4} \sqrt{\pi} \over 2\sqrt[4]{ab}}}$. Plugging this back in says that the original integral is
$$ e^{i \pi \over 4} {1 \over 2} \sqrt{\pi \over b} e^{2i \sqrt{ab}}$$
This is the answer. There are no convergence issues in the above; since the Fresnel integral converges the argument above is readily made rigorous, and at any rate one can use stationary phase a.k.a. integration by parts to directly show the original improper integral converges in the usual sense.
A: We can evaluate the following integral in a closed form, but under the conditions $Re(\alpha) < 0$ and $Re(\beta <0)$ 
$$ \int_{0}^{\infty} e^{\alpha x^2 + \frac{\beta}{x^2}}dx = -{\frac {\,i\sqrt {\pi}{{\rm e}^{2\sqrt{\beta}\sqrt{\alpha}}}}{2\sqrt {\alpha}}}. $$
