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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 $\int_{-\infty}^\infty e^{-(ax^2+bx+c)}~dx=\sqrt{\dfrac{\pi}{a}}e^{\frac{b^2-4ac}{4a}}$ ,

$e^{as^2}=\dfrac{1}{\sqrt{4a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2}{4a}-sx}~dx$

$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2+4asx}{4a}}~dx$

$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2+4asx+4a^2s^2-4a^2s^2}{4a}}~dx$

$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{(x+2as)^2-4a^2s^2}{4a}}~dx$

$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2-4a^2s^2}{4a}}~dx$

$=\dfrac{1}{\sqrt{a\pi}}\int_0^\infty e^{-\frac{x^2-4a^2s^2}{4a}}~dx$

$=\dfrac{1}{\sqrt{a\pi}}\int_{-2as}^\infty e^{-\frac{(x+2as)^2-4a^2s^2}{4a}}~dx$

$=\dfrac{1}{\sqrt{a\pi}}\int_{-2as}^\infty e^{-\frac{x^2}{4a}-sx}~dx$

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    @Fabian, does it mean that $\mathcal{B}^{-1}_{s\to x}\{e^{as^2+bs}\}$ can be found by the formulae $\int_{-\infty}^\infty e^{-(ax^2+bx+c)}~dx=\sqrt{\dfrac{\pi}{a}}e^{\frac{b^2-4ac}{4a}}$ and $\int_{-\infty}^\infty e^{-ax^2}\cos ux~dx=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^{-\frac{u^2}{4a}}$ only when the region of convergence of $s$ is assumed to be $(-\infty,\infty)$ ?2012-09-11
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    @Fabian, I really really want to find the inverse one-sided Laplace transform of the strange functions, especially some cases of the entire functions e.g. $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , $\mathcal{L}^{-1}_{s\to x}\{\sin s\}$ , $\mathcal{L}^{-1}_{s\to x}\{\cos s\}$ , etc. But the references about finding the inverse one-sided Laplace transform from first principle only focus on the some cases of the meromorphic functions but not the some cases of the entire functions. Dose it mean that inverse one-sided Laplace transform of the entire functions are especially hard to find?2012-09-11
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    @Fabian, how to practically apply http://en.wikipedia.org/wiki/Inverse_Laplace_transform#Mellin.27s_inverse_formula in e.g. $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , $\mathcal{L}^{-1}_{s\to x}\{\sin s\}$ , $\mathcal{L}^{-1}_{s\to x}\{\cos s\}$ , etc. and convert them in terms of integrals with real lower limit and upper limit? Or these are impossible and should only do sadly by http://en.wikipedia.org/wiki/Post%27s_inversion_formula?2012-09-12

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