I've seen in books like "Mathematical Handbook of Formulas and tables" by Spiegel et al. (3rd. edition) that
$$ \int_0^\infty dx\ e^{-ax^2 - b/x^2} = \frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}} \tag1$$
Nevertheless, in order to solve a Physics problem I only get the right solution if I change $\sqrt{\frac{\pi}{a}} \leftrightarrow \sqrt{\frac{\pi}{b}}$
Is there any chance of that formula to be an erratum? I've already computed with Wolfram-Apha and gives the same result as the book, but I'm pretty sure of not being wrong with the Physics and previous calculations.
Moreover, this integral appears when you try to compute
$$ I = \int_{\mathbb{R}^3}d^3p\ \frac{e^{i\vec{p}\vec{x}}}{(2\pi)^3(\vec{p}^2 + m^2)} = \frac{e^{-mx}}{4\pi x}, \quad x = |\vec{x}| \tag2$$ By using
$$ \frac{1}{\vec{p}^2 + m^2} = \int_0^\infty dy\ e^{-y(m^2 + \vec{p}^{\ 2})} \tag3$$
So the only way Eq. (2) is right is if Eq. (1) needs the change I suggest or if I'm using it wrongly.
ADDENDUM Let's see this last thing explicitly:
Introducing Eq. (3) in Eq. (2) renders,
$$ I = \int_0^\infty dy\int_{\mathbb{R}^3}\frac{d^3p}{(2\pi)^3}\ e^{-y\cdot m^2}e^{-y\vec{p}^{\ 2} + i\vec{p}\vec{x}} $$
Now, apply
$$ \int_{\mathbb{R}^3}\frac{d^3p}{(2\pi)^{3/2}}e^{-y\vec{p}^{\ 2} + i\vec{p}\vec{x}} = \frac{1}{\sqrt{2y}}e^{-x^2/(4y)} $$
So there is just one remaining integral in $y$ inside $I$, let's call it $I_y$, that you can solve with Eq. (1) (without my suggestion and using the change of variable $y = \alpha^2$):
$$ I_y = \int_0^\infty dy \frac{e^{-y\cdot m^2 - x^2/(4y)}}{\sqrt{y}} = \frac{\sqrt{\pi}}{m}e^{-mx} $$
And the problem, the reasion why you don't get Eq. (2) is due to the factor $1/m$ that should be $2/x$ to obtain Eq. (2)
It's convenient to put $a=r^2$ and $b=s^2$ and consider $$I=e^{2rs}\int_0^\infty\exp(-r^2x^2-s^2/x^2)\,dx =\int_0^\infty\exp(-(rx-s/x)^2)\,dx.$$ Letting $x'=s/(rx)$ gives $$I=\frac sr\int_0^\infty\exp(-(s/x'-rx')^2)\,\frac{dx'}{x'^2}$$ and so $$2I=\int_0^\infty\left(1+\frac{s}{rx^2}\right)\exp(-(rx-s/x)^2)\,dx.$$ Now define $y=rx-s/x$, to get $$2I=\frac1r\int_{-\infty}^\infty \exp(-y^2)\,dy=\frac{\sqrt\pi}{r}.$$