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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 $Re(x)<-1$?

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  • $\begingroup$ Extending is trivial, as we have $H_{x-1}=H_x-\frac1x$. $\endgroup$ – Simply Beautiful Art Jan 10 at 3:12
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Harmonic Numbers can also be extended to all $z\in\mathbb{C}$, except the negative integers, as $$H(z)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z}\right)$$

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