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I would like to compute the inverse Laplace of

$$F(s) = e^{-s^{\alpha}}$$

For the case $\alpha = \frac{1}{2}$ I've found the following on this website:

Compute the inverse Laplace transform of $e^{-\sqrt{z}}$

However, the accepted solution uses completing the square as one of its final steps, which is not possible fo $\alpha \neq \frac{1}{2}$, so I could no simply adjust the solution to solve my problem.

I come to the point where I have to solve:

$$f(t)=\int_{0}^{\infty}\frac{2}{\pi}ue^{u^2t}sin(u^{2\alpha}) du $$

to get the inverse function.

Any help is appreciated.

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  • $\begingroup$ What sort of thing are you expecting to get for this? There is certainly no expression in terms of elementary functions for this. There's some hope if $\alpha$ is a positive reciprocal integer, but not much beyond that. $\endgroup$ – Eric Towers Apr 6 '19 at 19:04
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    $\begingroup$ $\mathcal{L}_s^{-1}\left[\exp \left(-s^{\alpha }\right)\right](t)=\sum _{j=0}^{\infty } \frac{(-1)^j t^{-1-j \alpha }}{j! \Gamma (-j \alpha )}$ with CAS for:$\alpha =\frac{2}{3}$ is:$\frac{2 e^{-\frac{2}{27 t^2}} \text{Ai}\left(\frac{1}{3 \sqrt[3]{3} t^{4/3}}\right)}{3 \sqrt[3]{3} t^{7/3}}-\frac{2 e^{-\frac{2}{27 t^2}} \text{Ai}'\left(\frac{1}{3 \sqrt[3]{3} t^{4/3}}\right)}{3^{2/3} t^{5/3}}$ where:$\text{Ai}(t)$ and $ \text{Ai}'(t)$ is the Airy function and derivative of the Airy function. $\endgroup$ – Mariusz Iwaniuk Apr 6 '19 at 19:51
  • $\begingroup$ Maruisz, thank you showing this result. Which CAS did you use? I tried Maple and Wolfram with no succes. $\endgroup$ – pivu0 Apr 6 '19 at 20:00
  • $\begingroup$ I used Mathematica 11.3. $\endgroup$ – Mariusz Iwaniuk Apr 6 '19 at 21:08
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You are not going to integrate this in elementary terms.

One way to proceed, which requires some care in interpretationn is: $\mathrm{e}^{-s^\alpha} = \sum_{k=0}^\infty \frac{(-s^\alpha)^k}{k!}$, so \begin{align*} \mathscr{L}^{-1} \left\{ \mathrm{e}^{-s^\alpha} \right\}(t) &= \mathscr{L}^{-1} \left\{ \sum_{k=0}^\infty \frac{(-s^\alpha)^k}{k!} \right\}(t) \\ &\overset{?}{=} \sum_{k=0}^\infty \mathscr{L}^{-1} \left\{ \frac{(-s^\alpha)^k}{k!} \right\}(t) \\ &= \sum_{k=0}^\infty \frac{(-1)^k t^{-k\alpha - 1}}{k! \Gamma(-k \alpha)} \text{,} \end{align*} where the "?" is because we could easily have lost equality due to swapping limits. And we did: that denominator is trouble in the first term of the sum ... and that problem is replicated into every term if $\alpha$ is a nonnegative integer.

We can sidestep the problem in the first term. Let $$ f(\alpha) = \lim_{k \rightarrow 0} \left( \frac{(-1)^k t^{-k \alpha - 1}}{k! \Gamma(-k \alpha)} \right) + \sum_{k=1}^\infty \frac{(-1)^k t^{-k \alpha - 1}}{k! \Gamma(-k \alpha)} \text{,} $$ wherever that series converges. Then, where $\mathrm{Ai}(x)$ is the Airy function, $J_1(x)$ is the Bessel function of the first kind, and ${}_0\mathrm{F}_n(q_1, q_2, \dots, q_n; z)$ is a generalized hypergeometric function, \begin{align*} f(1/2) &= \frac{\mathrm{e}^{-1/(4t)}}{2 \sqrt{\pi}t^{3/2}} \\ f(1/3) &= 3^{-1/3}t^{-4/3}\mathrm{Ai}(3^{-1/3}t^{-1/3}) \\ f(2/3) &= \frac{2}{3^{4/3}}t^{-7/3}\mathrm{e}^{-2/(27t^2)} \left( \mathrm{Ai}(3^{-4/3}t^{-4/3}) - 3^{2/3} t^{2/3} \mathrm{Ai}'(3^{-4/3}t^{-4/3}) \right) \\ f(1/4) &= \frac{-{}_0\mathrm{F}_2(1/2, 3/4; -1/(256t))}{t^{5/4}\Gamma(-1/4)} - \frac{{}_0\mathrm{F}_2(3/4,5/4;-1/(256t))}{4 \sqrt{\pi}t^{3/2}} - \frac{{}_0\mathrm{F}_2(5/4,3/2;-1/(256t))}{6 t^{7/4}\Gamma(-3/4)} \\ f(-1/2) &= \frac{-{}_0\mathrm{F}_2(1/2, 3/2; t/4)}{\sqrt{\pi t}} + \frac{1}{2}{}_0\mathrm{F}_2(3/2,3; t/4) \\ f(-1) &= -J_1(2\sqrt{t})/\sqrt{t} \\ &= -{}_0\mathrm{F}_1(2; -t) \\ f(-2) &= -t {}_0\mathrm{F}_2(3/2,2; -t^2/4) \\ f(-3) &= \frac{-t^2}{2} {}_0\mathrm{F}_3(4/3, 5/3,2; -t^3/27) \\ f(-n) &= \frac{-t^{n-1}}{(n-1)!} {}_0\mathrm{F}_n \left( \frac{n+1}{n}, \frac{n+2}{n}, \dots, \frac{2n}{n}; -t^n/(n^n) \right) \text{ for } n \in \mathbb{Z}, n < 0 \end{align*} and, for most of these, we've already left the realm of recognizable functions; they don't get more recognizable if the arguments get more interesting. For evaluation by approximation, I suppose it's convenient the definitional series for the ${}_0\mathrm{F}_n$ functions are rapidly converging.

I have an appointment. I'll think about what to do if $\alpha$ is a nonnegative integer.


Note that $\alpha = 0$ gives $\delta(t)/\mathrm{e}$, which has the inconvenience of not actually being a function. (I.e., there's no hope of avoiding a limit, or possibly only defining it weakly, in terms of some integral operator.) And this generically happens for $\alpha \in \mathbb{Z}$, $\alpha \geq 0$. It's not clear what sort of answer you are looking for when the best answer is a distribution.

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  • $\begingroup$ How about when non-integer $\alpha$ and $>1$ ? $\endgroup$ – doraemonpaul Apr 13 '19 at 3:36
  • $\begingroup$ @doraemonpaul : When $1 < \Re \alpha <3$, $5 < \Re \alpha < 7$, $\dots$, the integral in the definition of the inverse Laplace transform does not converge. (The integrand of the improper integral on the path $s = \gamma + \mathrm{i}y$ doesn't go to zero as $y$ goes to $\pm \infty$, so the integral does not exist in the usual sense. Since exponentials are entire, take $\gamma = 0$ to see this for yourself.) Outside those intervals, there is more to say. Working on that. $\endgroup$ – Eric Towers Apr 14 '19 at 20:33
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    $\begingroup$ @Cechco : "the linearity of inverse Laplace transform" requires some regularity, for instance that each of the inverse transforms exist, which fails here. I don't say that $f(\alpha)$ always equals the desired inverse Laplace transform. The desired inverse Laplace transform frequently does not exist (see prior comments), so cannot agree with $f$. Where we are now is "if there is an inverse transform for a choice of $\alpha$, provisionally, it is $f$." And that claim is based on analytic continuation around the singularity blocking the question-marked equality in the first display. $\endgroup$ – Eric Towers Aug 10 '20 at 19:04
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    $\begingroup$ @Cechco : Start with real analysis: measure theory, Lebesgue spaces, and function spaces. Study operator theory, generalize to fractional operators (example and another), then (try to) use completeness to get paths in operator space. $\endgroup$ – Eric Towers Aug 28 '20 at 19:38
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    $\begingroup$ @Cechco : You seem confused by "if there is an inverse transform for a choice of $\alpha$, provisionally, it is $f$." (Aug 10, 19:04, comment in reply to you) This does not mean that $f(\alpha)$ has the same domain as the inverse Laplace transform (along $s$) of $\mathrm{e}^{-s^\alpha}$. Again, $f$ is an analytic continuation of that transform applied to that family of functions. $\endgroup$ – Eric Towers Aug 30 '20 at 14:38
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This inverse Laplace can be expressed in terms of the Fox H function

$\mathcal{L}^{-1}[e^{-s^\alpha}]=\displaystyle \frac{1}{z}H_{1,1}^{1,0}\left[z^{-\alpha}\Bigg\vert\begin{matrix}(0,\alpha)\\(0,1)\end{matrix}\right]$.

See Mathai A M, Kishore R, Saxena R, and Hausbold H J, 2010,The H-Function, Theory and applications, Springer, New York, equation (2.23 or 2.28).

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  • $\begingroup$ Worth nothing that the $H$ function cannot give us distributions. For instance, in $H$'s definition, for $\alpha = 0$, the $\Gamma(0)$ term in the denominator crushes the integrand and $\mathscr{L}^{-1}(\mathrm{e}^{-1})(t) = \delta(t)/\mathrm{e}$ is projected onto the zero function. I'm not complaining about your answer; capturing the distributions produced by the inverse Laplace transform is not easy. $\endgroup$ – Eric Towers Apr 14 '19 at 16:16
  • $\begingroup$ Indeed I forgot to add the restriction that this holds for $\alpha>0$. Could you elaborate on 'H function cannot give us distributions'? Best regards $\endgroup$ – Yakari Dubois Apr 15 '19 at 23:44
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    $\begingroup$ For $\alpha = 1/2$, the integrand of the Fox $H$ function diverges exponentially rapidly as the point on the path $L$ ($T$ in Fox's 1961 paper) approaches $\pm \mathrm{i}\infty$ (or the endpoints of the other two contour options). Similar divergence occurs for $\alpha = 1$. Unlike for the inverse Laplace transform, the integral is not conditionally convergent, so there is no hope of getting the distribution $\delta(z-1)$ from $H$ for $\alpha = 1$. Similar divergence occurs for $\alpha = -1$. Are you sure there isn't a typo' in your formula? $\endgroup$ – Eric Towers Apr 16 '19 at 1:40

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