Generalization of Harmonic series and Euler-Mascheroni Constant It is like a two-part question, and if answers already exist on Stack, would you be kind enough to add links to it.
First is a generalization of the Harmonic series. I made three assumptions:


*

*For $n \in \mathbb{N}$, $h(n)=H_n \\$; 

*$h(1)=1 \\$;

*$h(z)=h(z-1)+ \dfrac{1}{z}$.


And I let $f(n)$ be $\int\limits_1^n h(z) dz$. At the end, I found $$\int\limits_1^n h(z) dz = \ln (n!) + (n-1) \cdot \int\limits_{0}^1 h(z) dz.$$
My problem is with $\int\limits_{0}^1 h(z) dz$. Is there a way to show that it is equal to $\lim\limits_{n \rightarrow \infty} H_n - \ln (n)$? Or how can I show that $\int\limits_{0}^1 h(z) dz$ is convergent and equal to some number between 0 and 1?
Thank you, in advance.
 A: Starting with the definition of the harmonic number 
$$H_z=\int_0^1\frac{1-x^z}{1-x} \ dx$$
we can write 
$$\int_0^1 H_z \ dz=\int_0^1\frac1{1-x}\left(\int_0^1(1-x^z)\ dz\right)\ dx\\=\int_0^1\frac1{1-x}\left(\frac{1-x+\ln x}{\ln x}\right)\ dx\\=\int_0^1\left(\frac1{\ln x}+\frac{1}{1-x}\right)\ dx\\=\gamma=\lim_{n\to \infty}\left(H_n-\ln n\right)$$
where $\gamma$ is the Euler constant
About the last limit, you can find variant proofs here.
A: Or we can start with the series representation of the harmonic number:
$$H_z=\sum_{k=1}^z\frac1k=\lim_{n\to \infty}\sum_{k=1}^n\left(\frac1k-\frac1{z+k}\right)$$
Then
$$\int_0^1 H_z dz=\lim_{n\to \infty}\sum_{k=1}^n\int_0^1\left(\frac1k-\frac1{z+k}\right)\ dz\\=\lim_{n\to \infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=1}^n(\ln(1+k)-\ln(k))\right)\\=\lim_{n\to \infty}\left(H_n-\ln(1+n)\right), \quad 1+n\to n\\=\lim_{n\to \infty}\left(H_{n-1}-\ln(n)\right)\\=\lim_{n\to \infty}\left(H_{n}-\frac1n-\ln(n)\right)\\=\lim_{n\to \infty}\left(H_{n}-\ln(n)\right)$$
Note that $\sum_{k=1}^n(\ln(1+k)-\ln(k))=\ln(1+n)$ follows from using the telescoping series.
