The Normal definition of Harmonic numbers with $ n \in \mathbb{N} $ is
$$ H_n = \sum_{k=1}^{n}\frac{1}{k} \tag{1}\label{eq1A} $$
This can be expanded to $ n \in \mathbb{C} $
$$ H_n = \psi_0(n+1) + \gamma \tag{2}\label{eq2A}$$
Where $\psi_0(n)$ is the $0$th degree Polygamma function, which is defined for complex values of n, and $\gamma$ is the Euler-Mascheroni constant.
So my question is there a general solution to $H_{ji}$ where $j \in \mathbb{N} $ and $i$ is the imaginary unit?
I considered in using the series formula of the Polygamma which is
$$ \psi_0(z+1) = -\gamma + \sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+z}) \tag{3}\label{eq3A}$$
Simplify the $(2)$ with $(3)$ we get
$$ H_{ji} = \sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+ji}) \tag{4}\label{eq4A}$$
But I don't know how to simplify this series.
I tried to use the function above and Wolfram Alpha just simplifies it back to $\psi_0(x+1) + \gamma $ so this type of method seems to just be dead end.
Another method I have considered is the integral representation of $\psi_0(z)$ which is
$$ \psi_0(z) = \int_0^\infty (e^{-t}-\frac{1}{(1+t)^z})\frac{dt}{t} \tag{5}\label{eq5A}$$
Which transforms our $(2)$ into
$$ H_{ji} = \int_0^\infty (e^{-t}-{(1+t)^{-ji+1}})\frac{dt}{t} + \gamma \tag{6}\label{eq6A}$$
Expanding the integral int $(6)$ we get
$$ H_{ji} = \int_0^\infty \frac{e^{-t}}{t}dt - \int_0^\infty \frac{(1+t)^{-ji+1}}{t}dt + \gamma \tag{7}\label{e71A}$$
Which is just
$$ H_{ji} = \Gamma(0) - \int_0^\infty \frac{(1+t)^{-ji+1}}{t}dt +\gamma \tag{8}\label{eq8A}$$
Where $ \Gamma(z) $ is the Gamma function.
Seeing that $ \lim_{z\to 0} \Gamma(z) \rightarrow \infty $ and that the second integral doesn't converge there has to be some type of manipulation for $\Gamma(0)$ and the second integral to get a value of $H_{ji}$.
You can also use the identity of $ \psi(z+1) = \psi(z)+\frac{1}{z} $ to obtain for the previous function as
$$H_{ji} = \Gamma(0) - \int_0^\infty \frac{1}{(1+t)^{ji}t}dt - \frac{i}{j} +\gamma\tag{9}\label{eq9A}$$
Letting $ 1+t = u $ we can see our $(9)$ will change to
$$H_{ji} = \Gamma(0) - \int_1^\infty \frac{1}{u^{ji}(u-1)}du - \frac{i}{j} +\gamma \tag{10}\label{eq10A}$$
Doing partial fraction decomposition of $ \frac{1}{u^{ji}(u-1)} = \frac{A}{u^{ji}}+\frac{B}{u-1} $ we see that
$$ A = -1 \\ B = 1^{1-ji} \tag{11}\label{eq11A}$$
Expanding our $(10)$ with $(11)$ we get
$$H_{ji} = \Gamma(0) - \int_1^\infty (\frac{1^{1-ji}}{u-1} - \frac{1}{u^{ji}})du - \frac{i}{j} +\gamma \tag{12}\label{eq12A}$$
Constructing $(12)$ into two integrals and simpifying we see find
$$ H_{ji} = \Gamma(0) - \int_1^\infty \frac{1}{u-1} du - \int_1^\infty \frac{1}{u^{ji}} du - \frac{i}{j} + \gamma \tag{13}\label{eq13A}$$
But it is obvious that these two integrals don't converge to any value, so it seems that the partial fraction decomposition is also a dead end.
Thanks to @AliShather, they noticed that the integral in $(8)$ is very closly related to the Beta function, where the beta function is
$$ B(x,y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \Re\{x,y\}>0 \tag{14}\label{eq14A}$$
Comparing this to the integral in $(8)$ we can see that $\Re\{x,y\} \ngtr 0$, but very close!
Is there any better way to solve this?
Thank you for your time and patience!
$$\int_0^\infty\frac{(1+t)^{ji-1}}{t}dt=\int_0^\infty\frac{t^{-1}}{(1+t)^{-ji+1}}dt$$
and beta function
$$\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}d=\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
– Ali Shadhar Dec 08 '19 at 03:42