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

Usually, it is easier to compute asymptotics for an integral than for a sum. For example, Stirling formula can be deduced from the integral form given by the Gamma function by using the Laplace's method.

Another somewhat more elementary example of the relationship between series and integrals is given by comparing a series with an integral in order to assert its convergence.

Answer 2

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 $\Re(x)>-1$.