Analytic continuation for Harmonic series

Question

It is known that $$H_x=\sum_{n=1}^x\frac1n$$Which is very well known because of its interesting behavior as $x\rightarrow\infty$. Now this sum is defined only for positive integers $x$ because it is at the top of the summand. However, in this link I found a telescoping series that becomes the Harmonic series. It looks like this: $$H_x=\sum_{n=1}^\infty\frac x{n(n+x)}$$ This sum is defined for all numbers except the negative integers (like the Gamma function apparently…). This sum has been written before on stackexchange (I don't have the link right now), but usually this integral is preferred: $$H_x=\int_0^1\frac{t^x-1}{t-1}dt$$ Which is defined only for positive $x$. Is there any reason why this integral is preferred over the sum?

Answer 1

The sum and integral are equivalent.

Note

$$\frac{1}{t-1}=-\sum\limits_{n=1}^\infty t^{\,n-1},\quad -1<t<1\tag{1}$$

so

$$\int\limits_0^1 \frac{t^x-1}{t-1}\,dt=-\int\limits_0^1 \left(t^x-1\right)\, \sum\limits_{n=1}^\infty t^{\,n-1}\,dt=-\sum\limits_{n=1}^\infty \int_\limits0^1 \left(t^x-1\right)\, t^{\,n-1}\,dt=\sum\limits_{n=1}^\infty \frac{x}{n\, (n+x)}\tag{2}$$

which is valid for $\Re(x)>-1$.