http://en.wikipedia.org/wiki/Integral_transform#Table_of_transforms claims than the integral form of inverse bilateral Laplace transform and inverse Laplace transform are both the same. But are they true in reality?
For $\mathcal{B}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , since http://en.wikipedia.org/wiki/Lists_of_integrals#Definite_integrals_lacking_closed-form_antiderivatives has a formula that $\int_{-\infty}^\infty e^{-(ax^2+bx+c)}~dx=\sqrt{\dfrac{\pi}{a}}e^{\frac{b^2-4ac}{4a}}$ and http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourCos3.pdf has a formula that $\int_{-\infty}^\infty e^{-ax^2}\cos ux~dx=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^{-\frac{u^2}{4a}}$ , it is clear that $\mathcal{B}^{-1}(e^{as^2+bs})(x)$ exist a close-form whenever $a<0$ or $a>0$ .
However, for $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , it is troublesome. Since http://eqworld.ipmnet.ru/en/auxiliary/inttrans/laplace3.pdf has a formula that $\int_0^\infty e^{-(ax^2+px)}~dx=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^{\frac{p^2}{4a}}\mathrm{erfc}\left(\dfrac{p}{2\sqrt{a}}\right)$ , it seems that $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ difficult to have close-form. The difficulty of $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ seems to be far away from $\mathcal{B}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , and make me doubt the accuracy in http://en.wikipedia.org/wiki/Integral_transform#Table_of_transforms.
So is it possible to express $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ in terms of integrals with real lower limit and upper limit?
HINTS:
By $\int_{-\infty}^\infty e^{-(ax^2+bx+c)}~dx=\sqrt{\dfrac{\pi}{a}}e^{\frac{b^2-4ac}{4a}}$ ,
$e^{as^2}=\dfrac{1}{\sqrt{4a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2}{4a}-sx}~dx$
$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2+4asx}{4a}}~dx$
$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2+4asx+4a^2s^2-4a^2s^2}{4a}}~dx$
$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{(x+2as)^2-4a^2s^2}{4a}}~dx$
$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2-4a^2s^2}{4a}}~dx$
$=\dfrac{1}{\sqrt{a\pi}}\int_0^\infty e^{-\frac{x^2-4a^2s^2}{4a}}~dx$
$=\dfrac{1}{\sqrt{a\pi}}\int_{-2as}^\infty e^{-\frac{(x+2as)^2-4a^2s^2}{4a}}~dx$
$=\dfrac{1}{\sqrt{a\pi}}\int_{-2as}^\infty e^{-\frac{x^2}{4a}-sx}~dx$