# $\mathcal{B}^{-1}_{s\to x}\{e^{as^2+bs}\}$ and $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , where $a\neq0$

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 $(-\infty,\infty)$ ? –  doraemonpaul Sep 11 '12 at 23:32
@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? –  doraemonpaul Sep 11 '12 at 23:54
@Fabian, how to practically apply en.wikipedia.org/wiki/… 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 en.wikipedia.org/wiki/Post%27s_inversion_formula? –  doraemonpaul Sep 12 '12 at 0:04

The difference comes about because of the different analytic structure of the function in the frequency space (and I guess there is some mistake on the Wiki-page).

The inverse of the one-sided Laplace transform is indeed given by the Bromwich integral which coincides with what is written on the page. Here, the real constant $c$ is large enough such that the contour is to the right of all singularities. The reason is that if the one-sided Laplace transform exists, it converges for all $s$ with $c\leq \mathop{\rm Re}s$. The constant $c$ depends on the function in time-domain. If $f(t) \in O(e^{c t})$ then $f(s)$ is analytic for real part larger than $c$.

On the other hand, the inverse of the two-sided Laplace transform the constant $c$ should be chosen differently from what is claimed in the sentence after the table. In general, the Laplace transformed function will be only converge and thus be analytic in a strip with $c_-\leq\mathop{\rm Re}s \leq c_+$. The inverse transform has to be integrated within this strip.

What is important to notice: given a function in the frequency domain, the inversion process might not be unique. One always need the function and the strip of analyticity (i.e., the region where the original Laplace transform did converge). For the one-sided Laplace transform this remark is unimportant as the strip is always to the right of the last singularity. But to invert the two-sided Laplace transform, one need to integrate in between two singularities and thus there might be multiple choices. So a Laplace transformed function always need to be given with the region of analyticity:

e.g. consider the functions $$f_1(t) = \begin{cases} 0, & t<0\\ e^{2t}-e^{-3t},& t> 0\end{cases}$$ and $$f_1(t) = \begin{cases} -e^{2t}, & t<0\\ -e^{-3t},& t> 0\end{cases}.$$ The two-sided Laplace transform of the two functions coincides! $$\mathcal{B} (f_1) =\mathcal{B} (f_2) = \frac{5}{s^2+s - 6}.$$ However, the region of convergence of the first function is $\mathop{\rm Re}s >2$. Whereas the second function converges on the strip with $-3 < \mathop{\rm Re}s < 2$ in the complex plane.

The inverse of the one-sided Laplace transform will of course be $f_1$ (as by definition of the transform does not care about $t<0$).

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