Evaluating $\int x!\left(\frac1x+\frac{1}{x-1}+...+\frac11\right)dx$ A friend of mine told me to evaluate the following integral....
$$\int  x!\left(\frac1x+\frac{1}{x-1}+...+\frac11\right)dx$$
I have two questions here 
$1)$ Does the integral even actually exist in terms of elementary functions?
This is because  $x!$ is defined for integer values..
$2)$If the integral exists , how to evaluate it?
Multiplying makes the expression even more complicated....So , I tried writing this as $e^{\ln x!+\ln\left(\frac1x+\frac{1}{x-1}+...+\frac11\right)}$
But I don't know how to proceed further? Any help would be appreciated.....
 A: Let $\psi(z) = \frac{\Gamma'(z+1)}{\Gamma(z+1)}$, then we can resolve your ambiguous notation with the following.
$$\psi(z) = -\gamma + \sum_{n=1}^\infty \frac{1}{k} - \frac{1}{z+k}$$
$$\psi(n+1) - \psi(1) = \sum_{j=1}^n \frac{1}{j}$$
so that naturally
$$\psi(x+1) - \psi(1) = \frac{1}{x} + \frac{1}{x-1} + ... + 1$$
Therefore the integral you want to evaluate is
$$\int_a^z \Gamma(x+1)(\psi(x+1) - \psi(1))\,dx$$
which becomes at best
$$\Gamma(z+1) - \Gamma(a+1) - \psi(1)\int_a^z \Gamma(x+1)\,dx$$
A: I will try to give this question an actual meaning: at the moment, there are definition issues in $\frac{1}{x}+\frac{1}{x-1}+\ldots+1$ if $x\not\in\mathbb{N}$. We may start from the Weierstrass product for the $\Gamma$ function:
$$ \Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n} \tag{1}$$
and by applying $\frac{d}{dz}\log(\cdot)$ to both sides we have:
$$ \psi(z+1)\stackrel{\text{def}}{=}\frac{\Gamma'(z+1)}{\Gamma(z+1)} = -\gamma+\sum_{n\geq 1}\left[\frac{1}{n}-\frac{1}{n+z}\right]\tag{2} $$
hence:
$$ \frac{d}{dz}(z!) = \Gamma'(z+1) = \Gamma(z+1)\psi(z+1) = z!\left[-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right)\right]\tag{3} $$
and if $z\in\mathbb{N}$ the RHS of $(3)$ equals $z!\left[-\gamma+H_z\right]$.
