What is the name of the function $D(a,x) = \frac{x^a e^{-x}}{\Gamma(a+1)}$? 
Does the function $\dfrac{x^a e^{-x}}{\Gamma(a+1)}$ have its own specific name?

Temme [1] introduced the function in (3.1)
$$D(a,x) = \frac{x^a e^{-x}}{\Gamma(a+1)}.$$ It is the dominant part in many representations of the incomplete gamma functions (his corresponding Algol function is named dax).
Earlier DiDonato and Morris [2] called it $R(a,x)$.
A closely related function is called  regularised_gamma_prefix in the Boost library.
I did not find a 'name' for this function. Are there more known references to it?

[1]  N.M. Temme, A Set of Algorithms for the Incomplete Gamma Functions, Probability
in the Engineering and Informational Sciences, 8 (1994), pp. 291-307. Available as
https://ir.cwi.nl/pub/10080/10080D.pdf
[2] A.R. DiDonato, A.H. Morris, Computation of the Incomplete Gamma Function
Ratios and their Inverse. ACM TOMS, Vol 12, No 4, Dec 1986, pp. 377-393.
 A: You may see it called the probability density function of a gamma distribution. That is the restriction of this function to the interval $[0,+\infty).$ This distribution is that of the sum of $a+1$ independent random variables each of which is exponentially distributed, so that the probability of its being greater than $x$ is $e^{-x}$ for $x\ge0.$
One more often sees it written as $\dfrac{x^{\alpha-1} e^{-x}}{\Gamma(\alpha)}.$ That way $\alpha$ is the number of exponentially distributed random variables that are added.
Among those probability distributions called gamma distributions one also finds instances in which a scale parameter is present, thus:
$$
\frac {\left( x/\mu \right)^{\alpha-1} e^{-x/\mu}} {\Gamma(\alpha)} \left( \frac {dx} \mu\right) \text{ for } x\ge0. \tag{$\mu$}
$$
Sometimes it is parametrized using the reciprocal of the scale parameter, $\lambda= \dfrac 1 \mu{:}$
$$
\frac{(\lambda x)^{\alpha-1} e^{-\lambda x}}{\Gamma(\alpha)} (\lambda \, dx). \tag{$\lambda$}
$$
It may be that $\alpha$ is not a positive integer, in which case the statement about the sum of $\alpha$ random variables may seem not to make sense. But suppose $X_1,X_2$ are independent random variables with
$$
\Pr(X_i\in A) = \int_A \frac{x^{\alpha_i-1} e^{-x}}{\Gamma(\alpha_i)} \, dx \quad \text{for measurable } A \subseteq [0,+\infty), \quad i=1,2.
$$
In that case, one has
$$
\Pr(X_1+X_2\in A) = \int_A \frac{x^{\alpha_1+\alpha_2-1} e^{-x}}{\Gamma(\alpha_1+\alpha_2)} \, dx \quad \text{for measurable } A \subseteq [0,+\infty).
$$
This last probability density function is the convolution of those corresponding to $\alpha_1$ and $\alpha_2.$ Thus putting $\alpha-1$ in the exponent rather than $\alpha$ makes this a convolution semigroup in which convolution of these functions corresponds to addition of $\alpha$s.
When it's written in the $(\lambda)$ form rather than the $\mu$ form and $x$ is time in a Poisson process, then $\lambda x$ is the average number of arrivals in a time interval of length $x.$ When it's written in the $(\mu)$ form, then $\mu$ is the average time until the next arrival.
