# Representation for harmonic series $H_n$, for $n<-1$

According to Wolfram Alpha, harmonic series $$H_x$$ has the following representation: $$H_x=\int_{0}^{1}\frac{-1+t^x}{-1+t}dt=\int_{0}^{\infty}\frac{1-e^{-xt}}{-1+e^t}dt,~Re(x)>-1$$

The corresponding graph for $$H_x$$ will then be

However, Wolfram Alpha also gives the graph for $$Re(x)<1$$

But either one of the two functions above is undefined for $$Re(x)\leq-1$$. According to which function does Wolfram Alpha graph the harmonic series $$H_x$$, for $$Re(x)<-1$$? Is here such a function that satisfies both the values for $$Re(x)>-1$$ and the values graphed by Wolfram Alpha when $$Re(x)<-1$$?

• Extending is trivial, as we have $H_{x-1}=H_x-\frac1x$. – Simply Beautiful Art Jan 10 at 3:12

Harmonic Numbers can also be extended to all $$z\in\mathbb{C}$$, except the negative integers, as $$H(z)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z}\right)$$