$$\int_0^m \frac{n^x}{\Gamma(x+1)}dx:n,m \in \mathbb{R}$$ I'm dubious as to whether there's a closed form for the above, if there is I'll be very happy. Otherwise:
- Is there a closed form for $$\int_0^\infty \frac{n^x}{\Gamma(x+1)}dx?$$
- Is there an asymptotic expansion for $\int_0^m \frac{n^x}{\Gamma(x+1)}dx$ as $n \to \infty$ with $m$ constant? By this I mean $$\int_0^m \frac{n^x}{\Gamma(x+1)}dx=\sum_{i=0}^{m-1}\frac{n^i}{i!}+O(\text{something})$$ , or something similar.
(Note that the first integral is essentially the integral of the continuous analogue of the Poisson distribution).
For $\int_0^m\dfrac{n^x}{\Gamma(x+1)}dx$ , by using the formula
We have $\int_0^m\dfrac{n^x}{\Gamma(x+1)}dx=m\sum\limits_{s=1}^\infty\sum\limits_{r=1}^{2^s-1}\dfrac{(-1)^{r+1}n^\frac{rm}{2^s}}{2^s\Gamma\left(\dfrac{rm}{2^s}+1\right)}$
For $\int_0^\infty\dfrac{n^x}{\Gamma(x+1)}dx$ , according to http://books.google.hu/books?id=Pl5I2ZSI6uAC&pg=PA262&lpg=PA262&dq=Frans%C3%A9n%E2%80%93Robinson%20constant#v=onepage&q=Frans%C3%A9n%E2%80%93Robinson%20constant&f=false,
We have $\int_0^\infty\dfrac{n^x}{\Gamma(x+1)}dx=e^n-\int_{-\infty}^\infty\dfrac{e^{-ne^y}}{y^2+\pi^2}dy$