Can this infinite summation be simplified? I encountered the following infinite summation $$\sum_{k=0,k\neq m}^{\infty}\frac{x^k}{(k-m)k!},x>0,$$ can it be simplified? Thanks!
 A: As already mentioned, a possible approach consists in multiplying the sum by $x^{-m}$ which leads to a sum easy to express on exponential form. But what follows with this method is rather arduous, because further integration involves an incomplete Gamma function with negative integer parameter.
In order to save time, we will use another method (a short-cut in fact), starting from the known formula of the incomplete Gamma function in case of negative integer parameter :
$$\Gamma(-m,z)=\frac{(-1)^m}{m!}\big(\psi(m+1)-\ln(z)\big)-\frac{1}{z^m} \sum_{k=0,k\neq m}^{\infty}\frac{(-z)^k}{(k-m)k!}$$
$$\sum_{k=0,k\neq m}^{\infty}\frac{(-z)^k}{(k-m)k!}=z^m\left( \frac{(-1)^m}{m!}\big(\psi(m+1)-\ln(z)\big)-\Gamma(-m,z) \right)$$
For $x$ positive variable : $z=-x$ so that :
$$\sum_{k=0,k\neq m}^{\infty}\frac{x^k}{(k-m)k!}=(-x)^m\left( \frac{(-1)^m}{m!}\big(\psi(m+1)-\ln(-x)\big)-\Gamma(-m,-x) \right)$$
$\ln(-x)$ and $\Gamma(-m,-x)$ are both complex. But, the imaginary part fades in the formula. So, in separating the real part, we can write :
$$\sum_{k=0,k\neq m}^{\infty}\frac{x^k}{(k-m)k!}=(-x)^m\left( \frac{(-1)^m}{m!}\big(\psi(m+1)-\ln(x)\big)-\operatorname{Re}\big(\Gamma(-m,-x)\big) \right)$$
In this formula, $\psi$ is the symbol for the digamma function, $\Gamma$ is the symbol for the Incomplete Gamma function and $\operatorname{Re}$ is the symbol for the real part (to be not confused with the modulus, of course).
The formula has been tested with numerical computations.
