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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.

pivu0
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  • 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. – Eric Towers Apr 06 '19 at 19:04
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    $\mathcal{L}_s^{-1}\left\exp \left(-s^{\alpha }\right)\right=\sum _{j=0}^{\infty } \frac{(-1)^j t^{-1-j \alpha }}{j! \Gamma (-j \alpha )}$ 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. – Mariusz Iwaniuk Apr 06 '19 at 19:51
  • Maruisz, thank you showing this result. Which CAS did you use? I tried Maple and Wolfram with no succes. – pivu0 Apr 06 '19 at 20:00
  • I used Mathematica 11.3. – Mariusz Iwaniuk Apr 06 '19 at 21:08

2 Answers2

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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.

Eric Towers
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  • How about when non-integer $\alpha$ and $>1$ ? – doraemonpaul Apr 13 '19 at 03:36
  • @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. – Eric Towers Apr 14 '19 at 20:33
  • @EricTowers I am sorry for the perhaps stupid question. Is the first term (i.e., when $k=0$) $\mathscr{L}^{-1}(1) = \delta(t)$? – Alex Aug 08 '20 at 22:06
  • @Cechco : There are four series here indexed over $k$, three of which have a term for $k = 0$. Since you do not say which series you are thinking of, I guess you are talking about the second line of the first display. $(-s^\alpha)^k$ is not the constantly $1$ function -- it is not $1$ at $s = 0$. (This discontinuity in $s$ renders inapplicable many familiar theorems about the (inverse) Laplace transform.) You should perhaps note the "?" over the equality and the immediately following discussion explaining that equality failed starting at the second line of the first display. – Eric Towers Aug 09 '20 at 20:05
  • @EricTowers Hi Eric, Thank you! It seems to me that due to the reason that you pointed out, the equation for which we should put a question mark might be $\mathscr{L}^{-1}\left{ \frac{(-s^\alpha)^0}{0!} \right} \stackrel{?}{=} \frac{t^{-1}}{0!\Gamma(0)}$. Regarding the current equation with a question mark, one seems to be able to apply the linearity of inverse Laplace transform to the first term and all remaining terms as a whole. – Alex Aug 10 '20 at 18:30
  • @EricTowers I am sorry I have another question about the equation for $f(\alpha)$. $ \frac{(-1)^k t^{-k\alpha-1}}{k! \Gamma(-k\alpha)}$ is in trouble when $k=0$ because if $t=0$, both the numerator and denominator are $\infty$ (since $\Gamma(0)=\infty$). Am I right? And I couldn't get how you showed that $f(\alpha)$ equals to the desired inverse Laplace transform. – Alex Aug 10 '20 at 18:38
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    @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. – Eric Towers Aug 10 '20 at 19:04
  • @EricTowers Thank you so much for your detailed explanation! Q1: I am sorry I didn't quite get the meaning of the last sentence "claim is based on analytic continuation around the singularity blocking the question-marked equality in the first display". – Alex Aug 27 '20 at 06:16
  • Q2: I checked $f(1/3)$ and $f(1/4)$. $f(1/3)$ is continuous on $t\in [0,\infty)$. However, it seems to me that the limit of $f(1/4)$ at $t=0$ does not exist. I have a possible explanation for it in my mind but I am not sure of it. Maybe the inverse Laplace transform (ILT) of the first term $1$ is $\delta$ function. The ILT of the remaining terms also has a $\delta$ function but these $\delta$ functions cancel out in the sum. – Alex Aug 27 '20 at 06:18
  • Q3: Regarding the first term (i.e., $(-s^\alpha)^k$ when $k=0$), I think it is constantly 1 because the Taylor series says $e^x = 1+x+\frac{1}{2}x^2 +\dots$. Although the first term can be written as $x^0$, it is constantly 1 no matter what $x$ is. If it is constantly $1$, then the ILT of the first term is $\delta$ function. That's why I was confused by Q2 above. I am sorry that my expertise on complex analysis is limited and thank you again for your great answer! – Alex Aug 27 '20 at 06:23
  • @Cechco : Q1: From your Q3, "my expertise on complex analysis is limited". It is hopeless to attempt to explain analytic continuation in a comment. As a very simple example of what can (and is) happening -- the result of any integral (including the Laplace transform) is always continuous (see https://math.stackexchange.com/questions/429769/is-an-integral-always-continuous for the argument for Riemann and Lebesgue integrable functions). The result of summing an infinite collection of functions need not be continuous. (cont...) – Eric Towers Aug 27 '20 at 15:40
  • @Cechco : ... (See https://math.stackexchange.com/questions/2899174/sum-of-a-infinite-number-of-continuous-functions-on-a-set-may-not-be-continuous . Chinnapparaj R's answer there is an example of the singular behaviour you seem to be hung up on in your Q3.) Q2: You seem to think the only possibly singularities are delta functions, but, for example, $1/t$ is not a delta function. Q3: "Although the first term can be written as $x^0$, it is constantly $1$ no matter what $x$ is." is false. (cont...) – Eric Towers Aug 27 '20 at 15:41
  • @Cechco : You have already been told that this is false when $x = 0$ in my comment on Aug 9 at 20:05. There is no power of $0$ that is $1$. The $s = 0$, $k = 0$ term does not have numerator $1$. – Eric Towers Aug 27 '20 at 15:41
  • @EricTowers Thank you for your wonderful explanation! About Q1, I have ideas about analytic continuation and had experience computing it. If, for example, $0<\alpha<1/2$, $s=0$ is a branch point of $e^{-s^\alpha}$ and is the only singularity. I didn't quite get how your solution goes around this singularity. – Alex Aug 27 '20 at 16:52
  • @Cechco : The singularity is in the space of operators acting on functions. When you reverse the order of operators, you take two paths through that space. If the two paths (by analogy with complex analysis) "enclose a singularity in that space", their values need not agree. This means that the $\overset{?}{=}$ may or may not be an actual equality, depending on where the function we start at sits in relation to these operator singularities. – Eric Towers Aug 27 '20 at 16:57
  • @EricTowers About Q2, what confused me was that $f(1/3)$ is continuous on $[0,\infty)$ but $f(1/4)$ is not. What happens between $\alpha=1/3$ and $\alpha=1/4$? I thought $f(1/4)$ should also be continuous on $[0,\infty)$ (which might be incorrect, but I am not sure). – Alex Aug 27 '20 at 16:58
  • @Cechco : When $\alpha$ is not an integer, we have a problem of selection of branch, with a branch point at $0$. If $\alpha$ is rational, that point might be a regular branch point or it might be a singular branch point. I don't know of any easy way to predict which it will be. Why are you expecting "nice" behaviour there? – Eric Towers Aug 27 '20 at 17:02
  • @EricTowers I re-computed them and you are totally right. $f(1/3)$ and $f(1/9)$ have a right limit at $t=0$ but $f(1/4)$, $f(2/5)$ and $f(1/6)$ do not have a right limit at 0. – Alex Aug 27 '20 at 19:44
  • @EricTowers "The singularity is in the space of operators acting on functions." This is what I didn't know before. May I have some references that I could learn from? – Alex Aug 27 '20 at 22:44
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    @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. – Eric Towers Aug 28 '20 at 19:38
  • @EricTowers Theorem 28.2 of Gustav Doetsch's Introduction to the Theory and Application of the Laplace Transformation (see page 187) seems to imply that $f(\alpha)$ is continuous on $t\in \mathbb{R}$ if $0<\alpha<1$ (moreover, it says $f(\alpha)=0$ for $t<0$). Doetsch presented $e^{-s^\alpha}$ ($0<\alpha<1$) as an example right after Theorem 28.2 as an application of the theorem. – Alex Aug 30 '20 at 04:10
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    @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. – Eric Towers Aug 30 '20 at 14:38
  • @EricTowers Thank you for your clarification. It seems that I understand it now. In principle, one cannot invert $s^{\alpha k}$ (one can only get the inverse Laplace transform of $1/s^{v}$ with $\Re v>0$). However, if we pretend that it is true and do that anyway, we get the inverse transform as long as it exists for a choice of $\alpha$. This may be because the Laplace transform is analytic (however, I am a little confused here since we are talking about the inverse transform [Ch6 of Doetsch's book]). Is it analytic as well?). – Alex Aug 30 '20 at 18:58
  • @EricTowers How do you verify that $f(\alpha)$ is indeed the inverse Laplace transform when such an inverse transform exists, maybe by performing the Laplace transform? – Alex Aug 30 '20 at 18:59
  • @EricTowers Regarding my comment 14 hours ago (https://math.stackexchange.com/questions/3177322/inverse-laplace-transform-of-e-s-alpha/3177434?noredirect=1#comment7849203_3177434), I would like to provide a reference that studied this problem. According to Doetsch's book, such an inverse transform does not exist if $\alpha\ge 1$ (p.142). Moreover, this book discussed the continuity problem that I mentioned on Aug 27 when $\alpha\in (0,1)$. According to Theorem 28.2, if $\alpha\in (0,1)$, the inverse transform exists and is continuous at $0$. – Alex Aug 30 '20 at 19:08
  • @EricTowers However, in my comment on Aug 27, I wrote $f(1/4)$, $f(2/5)$, and $f(1/6)$ do not have a right limit at $0$. Mathematica gave me the right limit (it said the right limit was ComplexInfinity). It clearly contradicted Doetsch's Theorem 28.2. So I suspect that Mathematica has a bug here (which made me quite confused). I wrote my comment 14 hours ago in order to correct my previous comment if it was wrong and am looking forward to hearing your feedback regarding the right continuity at $0$ when $0<\alpha<1$. – Alex Aug 30 '20 at 19:12
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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).

Yakari Dubois
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  • 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. – Eric Towers Apr 14 '19 at 16:16
  • 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

     – Yakari Dubois Apr 15 '19 at 23:44
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    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? – Eric Towers Apr 16 '19 at 01:40