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Now since the sum $$ \sum_{n=0}^\infty \frac{x^n}{n!},\quad x\in\Bbb R, $$ does have some relatively nice properties, is the same true for its analogues integral? If we take the gamma function to be a generalisation of the factorial with $\Gamma(n+1) = n!$, an obvious analogues integral formula would be $$ \int_0^\infty \frac{x^t}{\Gamma(t+1)}\text dt,\quad x\ge 0. $$ Does this integral have any similar, nice properties?

Harry Peter
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Gaussler
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    huh. never thought about it! good question +1 – Bennett Gardiner Jun 01 '14 at 07:32
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    First of all, it should perhaps be noted that it does converge because of the integral criterion for real infinite series. – Gaussler Jun 01 '14 at 07:35
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    There is [something similar in another question](http://math.stackexchange.com/a/741634/85343). – Felix Marin Jun 01 '14 at 07:48
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    Yes, it does deal with a special case. But I don't think the general case follows from that. – Gaussler Jun 01 '14 at 08:18
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    Am I doing something dumb here? Let's try differentiating our integral. Call $$ f(x) := \int_0^{\infty} \frac{x^t}{\Gamma(t+1)} \ \mathrm{d}t $$ $$ \frac{\mathrm{d}f(x)}{\mathrm{d}x} = \int_0^{\infty} \frac{tx^{t-1}}{\Gamma(t+1)} \ \mathrm{d}t = \int_0^{\infty} \frac{x^{t-1}}{\Gamma(t)} \ \mathrm{d}t $$ due to the reflection formula for the Gamma function. Making a substitution of $u=t-1$ gives $$ \int_{-1}^{\infty} \frac{x^{u}}{\Gamma(u+1)} \ \mathrm{d}t $$ So that $$ \frac{\mathrm{d}f(x)}{\mathrm{d}x} = f(x) + \int_{-1}^{0} \frac{x^{u}}{\Gamma(u+1)} \ \mathrm{d}t $$ – Bennett Gardiner Jun 04 '14 at 04:18
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    @BennettGardiner definitely not dumb; indeed there is some analogue to the properties of the exponential function. :-) – Gaussler Mar 17 '15 at 15:54
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    Forgot to change the $t$ to a $u$ there. I wonder what you can say about that last integral given those bounds. – Bennett Gardiner Mar 18 '15 at 10:51

1 Answers1

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The integral can be rephrased, using the transformation $x=e^x$ in the same spirit as the general analogy of Taylor series to Laplace transforms, as $$ \int_0^\infty \frac{x^t \mathrm dt}{\Gamma(t+1)} =\int_0^\infty \frac{e^{st}\mathrm dt}{\Gamma(t+1)}. $$ This integral is known as the nu function, denoted $\nu(x)=\nu(e^s)$. Wikipedia and MathWorld have the same definition, but don't offer much detail. For more information, the best place to go is probably Erdelyi et al., Higher Trascendental Functions vol. 3, p.217, §18.3 (where chapter 18 is just 'miscellaneous functions'). Quoting from there:

[The function $\nu(x)$] was encountered by Volterra in his theory of convolution-logarithms (Volterra 1916 Chapter VI, Volterra and Péres 1924, Chapter X) [...]. These functions also occur in connection with operational calculus, appear in an inversion formula of the Laplace transformation, and are of interest in connection with certain integral equations.

Erdelyi et al. show that $$ \nu(x)=\begin{cases} e^x+O(|x|^{-N}) & |\arg(x)|\leq\pi/2 \\ O(|x|^{-N}) & \pi/2<|\arg(x)|\leq\pi \end{cases} $$ for any integer $N$, and that apart from being a simple(ish) Laplace transform itself, $\nu(x)$ has a simple Laplace transform, $$ \int_0^\infty e^{-st}\nu(t)\mathrm dt=\frac{1}{s\log(s)} \quad\operatorname{Re}(s)>1. $$ Both of these properties are (apparently) relevant for the use of $\nu(x)$ in operational calculus; Erdelyi et al. give references from the 1940s but presumably the field has moved on since then.

Regarding the relation to integral equations, the nu function obeys the rather pleasing equation $$ \int_0^\infty \exp\left(-\frac{x^2}{4y}\right)\nu(x)\mathrm dx =\frac{\sqrt{\pi y}}{2}\nu(y), $$ i.e. it is an eigenfunction of the integral kernel $y^{-1/2}\exp\left(-\frac{x^2}{4y}\right)$, solving an integral equation which "gives all characteristic functions which, in a certain sense, are of regular growth", though it is apparently a solution to other interesting equations of similar form.

For further details, see the references in Erdelyi et al.

E.P.
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