# An analogue of the exponential function by replacing infinite series with improper integral

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?

• for small $x$ the integral vanishes as $1/\log x$. – Carlo Beenakker Jan 29 at 12:50
• 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 $x^2/2$? – Gerry Myerson Jan 30 at 5:47
• @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. – Captain Emacs Jan 30 at 15:16
• This function was discussed on math.stackexchange some years back ... link – user26872 Feb 2 at 21:44
• @user26872 thank you for the link – Ali Taghavi Feb 2 at 21:49

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 $$C=\int_0^\infty\frac{dt}{\Gamma(t)}= 2.8077702420285...$$

• Thank you very much for your very interesting answer and the references you provided in the answer. – Ali Taghavi Jan 30 at 10:01

(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 $$\alpha$$).

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