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For every positive real number $x$ we define $$E(x)= \int_0^{\infty} x^t/t!\,\mathrm dt$$ where $t!=\Gamma(t+1)$. This is motivated by classical exponential function.

Is this function well defined (the problem of convergence)? Is there a real analytic extention of $E$ to all real numbers? What about a holomorphic extention to complex numbers? How can we compare $E(x+y)$ with $E(x)$ and $E(y)$? What kind of differential equation can be satisfied by $E$? Is $E$ one to one on real numbers?What can be said about its possible inverse?

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    $\begingroup$ for small $x$ the integral vanishes as $1/\log x$. $\endgroup$ – Carlo Beenakker Jan 29 at 12:50
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    $\begingroup$ In what sense is this a "generalization of the exponential function"? I understand that one is of the form $\sum_nf(n,x)$ while the other is $\int f(t,x)\,dt$, but that doesn't really make it a generalization, does it? Is $(n^2+n)/2$ a generalization of $x^2/2$? $\endgroup$ – Gerry Myerson Jan 30 at 5:47
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    $\begingroup$ @GerryMyerson Yes, in a way it is. Pre-Fermat's (as far as I know) approaches to integration of polynomials rely on summing up n^k over n and thus closed summation formulas (n is later taken to infinity). So, the term generalisation is very appropriate, at least under a historical perspective. $\endgroup$ – Captain Emacs Jan 30 at 15:16
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    $\begingroup$ This function was discussed on math.stackexchange some years back ... link $\endgroup$ – user26872 Feb 2 at 21:44
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    $\begingroup$ @user26872 thank you for the link $\endgroup$ – Ali Taghavi Feb 2 at 21:49
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This is particular case of a classic integral studied by Ramanujan. See Chapter 11 in Hardy's book, "Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work", where it is shown that $$ \int_{-\xi}^\infty\frac{x^t}{\Gamma(1+t)}\,dt+\int_0^\infty t^{\xi-1}e^{-xt}\left(\cos\pi \xi-\frac{\sin\pi \xi}{\pi}\ln t\right)\frac{dt}{\pi^2+\ln^2t}=e^x, \quad (x\ge 0, \xi\ge 0). $$ From this it follows that your integral can be represented as an integral of elementary functions as follows $$ \int_0^\infty\frac{x^t}{\Gamma(1+t)}\,dt=e^x-\int_0^\infty \frac{e^{-xt}\,dt}{t(\pi^2+\ln^2t)},\quad (x\ge 0). $$

Also, see Fransen-Robinson constant $$ C=\int_0^\infty\frac{dt}{\Gamma(t)}= 2.8077702420285... $$

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  • $\begingroup$ Thank you very much for your very interesting answer and the references you provided in the answer. $\endgroup$ – Ali Taghavi Jan 30 at 10:01
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(Some obvious properties of $E$; too long for a comment, though).

The holomorphic extension of $E$ to $\mathbb{C} \setminus (-\infty, 0]$ (in fact, to the entire Riemann surface of the complex logarithm) is given by $$E(x) = \int_0^\infty \frac{\exp(t \log x)}{\Gamma(t+1)}\, dt,$$ where $\log$ denotes the principal branch of the complex logarithm. This follows from a standard application of Morera's theorem, involving Fubini's theorem, the estimate $$|\exp(t \log x)| = \exp(t \log |x|) = |x|^t \le a^t + b^t$$ when $a \le |x| \le b$, and integrability of $(a^t + b^t) / \Gamma(t + 1)$ over $(0, \infty)$.


In particular, for $x > 0$ we have $\log(-x + 0 i) = \log x + i \pi$, and hence $$ E(-x+0i) = \int_0^\infty e^{i \pi t} \frac{x^t}{\Gamma(t+1)} \, dt $$ is not real-valued in any neighbourhood of $0$. Thus, there is no real-analytic extension of $E$ to $(-\epsilon, \infty)$ (as already follows from Carlo Beenakker's comment).


By dominated convergence theorem, the integral can be differentiated under the integral sign, so $$E^{(n)}(x) = \int_0^\infty \frac{x^{t - n}}{\Gamma(t+1 - n)}\, dt = \int_{-n}^\infty \frac{x^t}{\Gamma(t+1)}\, dt .$$ This does not seem to lead to any interesting differential equation.


Since $E'(x) > 0$, clearly $E$ is increasing on $(0, \infty)$, with $E(0) = 0$ and $E(\infty) = \infty$.


One can easily find the Laplace transform of $E(x)$: when $\operatorname{Re} \xi > 1$, we have $$ \int_0^\infty e^{-\xi x} E(x) dx = \int_0^\infty \frac{1}{\xi^{t + 1}} \, dt = \frac{1}{\xi \log \xi} . $$


(EDIT: This was meant to be an extended comment only, but since it has received a number of upvotes, let me add further remarks, inspired by Nemo's answer.)

The Laplace transform $\mathcal{L} E$ of $E$ has a simple pole at $\xi = 1$ with residue $1$, and a branch cut along $(-\infty, 0]$. Since it decays (barely) sufficiently fast at infinity, one can (carefully) write the usual inversion formula and then deform the contour of integration to the Hankel contour to find that $$ E(x) = e^x - \frac{1}{\pi} \int_0^\infty e^{-t x} \operatorname{Im} (\mathcal{L} E(-t + 0i)) dt .$$ This leads to the formula given in Nemo's answer: since $$\mathcal{L} E(-t + 0i) = -\frac{1}{t \log(-t + 0 i)} = -\frac{1}{t (\log t + i \pi)} \, ,$$ we obtain $$ E(x) = e^x - \int_0^\infty \frac{e^{-t x}}{t (\pi^2 + \log^2 t)} \, dt .$$ As a consequence, $e^x - E(x)$ is completely monotone, and $$ E(x) = e^x - \frac{1 + o(1)}{\log x} $$ as $x \to \infty$. Further terms can be obtained in a similar way.

The function $E(x)$ itself is the Mellin transform of $1 / \Gamma(t + 1)$. Thus, $1 / \Gamma(t + 1)$ can be written as the inverse Mellin transform: $$ \frac{1}{\Gamma(t + 1)} = \frac{1}{2 \pi i} \int_{c + i \mathbb{R}} t^{-1 - x} E(x) dx , $$ or, equivalently, $$ \frac{1}{\Gamma(t)} = \frac{1}{2 \pi i} \int_{c + i \mathbb{R}} t^{-x} E(x) dx . $$ The definition of $E(x)$ looks a little bit like Mellin–Barnes integral, but the contour is wrong.

Finally, the (fractional) integral of $E(x)$ of order $\alpha$ is given by $$ I_\alpha E(x) = \frac{1}{\Gamma(\alpha)} \int_0^x E(t) (x - t)^{\alpha - 1} dt = \int_0^\infty \frac{x^{t + \alpha}}{\Gamma(t + 1 + \alpha)} \, dt , $$ and so $$ I_\alpha E(x) = \int_\alpha^\infty \frac{x^t}{\Gamma(t + 1)} \, dt . $$ This agrees with the expression for the derivatives of $E$ (which correspond to negative integer $\alpha$).

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  • $\begingroup$ "...not real-valued in no neighbourhood of zero." I'm having trouble parsing this. $\endgroup$ – Gerry Myerson Jan 30 at 5:42
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    $\begingroup$ @Gerry Myerson: "not real-valued in any (real) neighborhood of zero". $\endgroup$ – The_Sympathizer Jan 30 at 8:01
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    $\begingroup$ @GerryMyerson: We don't need no education... Thanks! $\endgroup$ – Mateusz Kwaśnicki Jan 30 at 8:22
  • $\begingroup$ Thank you very much for you very interesting answer and the concepts you introduced in your answer. $\endgroup$ – Ali Taghavi Jan 30 at 10:03

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