# Integral $\int_0^\infty \exp(ia/x^2+ibx^2)dx$

Compute the integral:

$$\int_0^\infty \exp\left(\frac{ia}{x^2}+ibx^2\right)\,dx$$

for $a$, $b$ real and positive. I tried complex variables, but don't really know how to handle this.

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The integral does not seem to exist. – Did Apr 7 '13 at 13:36
It should exist; book claims it is $\sqrt{i \pi / 4b}\, e^{2i\sqrt{ab}}$ – quark1245 Apr 7 '13 at 13:44
Which book? In which sense does it claim convergence since the function to be integrated has modulus $1$ everywhere? As a limit of the integral from $0$ to $R$ when $R\to\infty$? – Did Apr 7 '13 at 13:46
Concerning real integrals there is the classical $$\int_0^\infty e^{-\frac a{x^2}-bx^2}dx=\sqrt{\frac{\pi}{4b}}e^{-2\sqrt{ab}}$$ (from the appendix of Feynman&Hibb's "Quantum Mechanics and Path Integrals" ($1965$ edition). By analytic continuation you may probably obtain your result... (with some ambiguity on the sign of the result...) – Raymond Manzoni Apr 7 '13 at 14:01
@quark1245, I've Feynman's book's appendix in front of me and I think you have a rather boogie-buggy typo: the book's integral has no $\,i\,$ at all in the exponential's argument, making it exactly what Raymond told you! If there was an $\,i\,$ then, as Did told you, the integral's usual meaning would diverge...In fact, Feynman's appendix has minus in the exponential argument ... – DonAntonio Apr 7 '13 at 14:32

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.

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You can basically do the same thing directly to the original integral; see my answer – Zarrax Apr 7 '13 at 17:34
Incredibly good answer, that's exactly the problem I was trying to do – quark1245 Apr 7 '13 at 20:00
Thanks @quark1245 (and I thought that $6-13$ could be appropriate...) and thanks Zarrax for the $i$ case. – Raymond Manzoni Apr 7 '13 at 21:12
(+1) Nice techniues. – Mhenni Benghorbal Apr 15 '13 at 0:52
I like it too @Mhenni ! A more general formula is $\ \displaystyle \int_0^\infty e^{-\frac{a^2}{x^n}-b^2\,x^n}dx=\frac 2n\left(\frac ab\right)^{\frac 1n}\ \operatorname{K}_{\frac 1n}(2\,ab)$ with $\operatorname{K}_n$ the modified Bessel function (cf $\,10.32.10$). For $n=2$ we got $\ \operatorname{K}_{\frac 12}(z)=\sqrt{\frac {\pi}2}\frac {e^{-z}}{\sqrt{z}}$, for other results ($n=3$ : Airy function and so on) see Wolfram functions. – Raymond Manzoni Apr 15 '13 at 20:32

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.

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+1 and very clear answer, but the other one had extra linked material – quark1245 Apr 7 '13 at 20:02

We can evaluate the following integral in a closed form, but under the conditions $Re(\alpha) < 0$ and $Re(\beta <0)$

$$\int_{0}^{\infty} e^{\alpha x^2 + \frac{\beta}{x^2}}dx = -{\frac {\,i\sqrt {\pi}{{\rm e}^{2\sqrt{\beta}\sqrt{\alpha}}}}{2\sqrt {\alpha}}}.$$

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