TITLE: Integral $\int_0^\infty \exp(ia/x^2+ibx^2)dx$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.