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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!

Dclrk
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    For your last integral, try $u=1/x$ then use beta function. – Ali Shadhar Dec 08 '19 at 02:41
  • Or you can use beta function for the integral with boundaries $(0,\infty)$ – Ali Shadhar Dec 08 '19 at 02:43
  • @AliShather Thank you for the reply! I tried to use partial fraction decomposition for the last integral, I will try your approach just give me a second. – Dclrk Dec 08 '19 at 02:54
  • Sure.. I hope it will work – Ali Shadhar Dec 08 '19 at 02:57
  • @AliShather When I substituted $ u = \frac{1}{x} $, I got $ \int_0^1 x^{ij-2}(\frac{1}{x}-1)^{-1}dx $ as my integral. Did I do something wrong or is this the result that you got to? – Dclrk Dec 08 '19 at 03:19
  • Ade you familiar with beta function? We have couple forms of it. It's on wiki if you would like to check. – Ali Shadhar Dec 08 '19 at 03:25
  • @AliShather I've done surface level Beta Function the one that's like $ B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1}dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} $. I have not done more in-depth definitions of it. – Dclrk Dec 08 '19 at 03:31
  • The restriction of beta function is that Re(x) and Re(y) >0 – Ali Shadhar Dec 08 '19 at 03:41
  • We can put your integral in beta function form,

    $$\int_0^\infty\frac{(1+t)^{ji-1}}{t}dt=\int_0^\infty\frac{t^{-1}}{(1+t)^{-ji+1}}dt$$

    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
  • @AliShather Oh, so our integral would become $\frac{\Gamma(0)\Gamma(-ji+1)}{\Gamma(-ji+1)}$. – Dclrk Dec 08 '19 at 03:49
  • @AliShather Ahh, so close yet so far. Is there no extension of the Beta function to $ x = 0 $? – Dclrk Dec 08 '19 at 03:57
  • by comparing we notice that $x=0$ and the restriction of $\Re{x}$ has to be $>0$ so I do not think we can apply beta function here but I know that the range of $x$ and $y$ can be extended in case we have the derivative of beta function. – Ali Shadhar Dec 08 '19 at 04:00
  • @Dcirk What is your question? Formula $(3)$ is a definition of the harmonic number valid for any complex $z$. Is shows that it is a meromorphic function with simple poles with residue 1 at the negative intergers. This is a complete description of the function which obviously cannot be simplified. – Dr. Wolfgang Hintze Dec 08 '19 at 07:22
  • @Dr.WolfgangHintze If $(3)$ is the general solution to $(H_{ji})$ why does it seem impossible to simplify the integral representation of $\psi_0(z)$ without encountering the pole of the Gamma function at $0$? – Dclrk Dec 08 '19 at 18:59
  • @ Dcirk The error is in $(4)$: here you split the integrand so that two divergent integrals result. The same would happen if you wold split the sum $(3)$ into two (divergent !) parts. Hence: you must keep these expressions together. I suggest, as an example, you try to calculate $H_i$, the harmonic number at the imaginary unit. – Dr. Wolfgang Hintze Dec 08 '19 at 19:12
  • @Ali editing lots of old questions in a short period pushes lots of new questions off the front page. Please don't do that. – Gerry Myerson Dec 09 '19 at 03:35
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    @Gerry Myerson sorry didnt know that. thanks for letting me know. at that time i didnt know how to choose the right tags and that's why i edited them yesterday and today. thanks again. much stuff here i am not familiar with. – Ali Shadhar Dec 09 '19 at 03:42

2 Answers2

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Let us consider the simplest case with an imaginary argument of the harmonic number, namely $H_i$. This can easily be generalized to the requested case $H_{ji}$.

The harmonic number can be defined for complex $z$ as

$$H_z = \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+z} \right)\tag{1}$$

Notice that you must not split the sum into two parts, because both sums

$$\sum_{k=1}^\infty \left( \frac{1}{k} \right),\sum_{k=1}^\infty \left( \frac{1}{k+z} \right) $$

are divergent.

Now for $z=i$ we have

$$H_i = \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+i} \right)\tag{2}$$

writing the summand as

$$ \frac{1}{k} - \frac{1}{k+i} = \frac{1}{k} - \frac{k-i}{(k+i)(k-i)} = \frac{1}{k} - \frac{k-i}{k^2+1}=\frac{1}{k} - \frac{k}{k^2+1}+i \frac{1}{k^2+1}\\=\frac{1}{k(k^2+1)} +i \frac{1}{k^2+1} $$

we get

$$H_i = \sum_{k=1}^\infty \left(\frac{1}{k(k^2+1)} +i \frac{1}{k^2+1}\right)\tag{3}$$

Now splitting is permitted because the two sums are convergent. In fact, the real and imaginary parts of $H_i$, $f$ and $g$, respectively, are

$$f = \sum_{k=1}^\infty \left(\frac{1}{k(k^2+1)}\right)\tag{4}$$

$$g = \sum_{k=1}^\infty \left( \frac{1}{k^2+1}\right)\tag{5}$$

I stop here for a while to let you calculate $f$ and $g$, i.e. find expressions which are not just real and imaginary part of $H_i$.

EDIT

Now, what can be said about $f$ and $g$?

$g$ has a closed form

$$g = \sum_{k=1}^\infty \left( \frac{1}{k^2+1}\right)= \frac{1}{2}\left(\sum_{k=-\infty}^\infty \left( \frac{1}{k^2+1}\right)-1 \right) \\= \frac{1}{2} (\pi \coth (\pi )-1) = \dfrac{\pi-1}{2}+\dfrac{\pi}{e^{2\pi}-1}\tag{6}$$

which has been given in several places in this forum, for instance here How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$ (notice that there the sum starts at $k=0$) and derived with complex contour integration here How to sum $\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}$?.

For $f$ I have not found a closed expression other than the information that $f$ is the real part of $H_i$. However, normally this would be considered a closed form as well.

Technically, the deeper reason for the different behaviour of $f$ and $g$ is that whereas $g$ can be written as a symmetric sum from $-\infty$ to $\infty$ which allows complex contour integration with a kernel $\pi \cot(\pi z)$, $f$ is a one-sided sum which has the kernel $H_{-z}$. The latter kernel then just brings us back to where we came from. Usage of contour integrals for infinite sums is described for example in chapter 2 of https://projecteuclid.org/download/pdf_1/euclid.em/1047674270

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In case the OP is curious about how to evaluate $f$ and $g$ in Dr. Wolfgang Hintze's solution above,


For $f$, we use the generalization $$H_a=\sum_{k=1}^\infty\frac{a}{k(k+a)}$$

so

$$ H_a+H_{-a}=\sum_{k=1}^\infty\frac{a}{k(k+a)}+\sum_{k=1}^\infty\frac{-a}{k(k-a)}=\sum_{k=1}^\infty\frac{-2a^2}{k(k^2-a^2)}$$

set $a=i$ we get

$$ H_i+H_{-i}=\sum_{k=1}^\infty\frac{2}{k(k^2+1)}$$

for $g$, we use the classical generalization

$$\sum_{k=1}^\infty\frac1{k^2+a^2}=\frac1{2a^2}\left(a\pi\coth(a\pi)-1\right)$$

$$\Longrightarrow g=\sum_{k=1}^\infty\frac{1}{k^2+1}=\frac12(\pi\coth(\pi)-1)$$

so

$$H_i=f+ig=\frac12H_i+\frac12H_{-i}+\frac i2(\pi\coth(\pi)-1)$$

or

$$H_i-H_{-i}=i(\pi\coth(\pi)-1)\tag1$$


By the way, we can quickly reach the result in $(1)$ if we use the identity

$$H_{-a}-H_{a-1}=\pi\cot(a\pi)$$

$$\Longrightarrow H_{-i}-H_{i-1}=-i\pi\coth(\pi)$$

plug $H_{i-1}=H_i-\frac1i=H_i+i$

we get

$$H_{-i}-H_i=-i(\pi\coth(\pi)-1)$$


Its obvious that $$\Re\{H_{-i}\}=\Re\{H_{i}\}$$

and

$$\Im\{H_{-i}\}=-\Im\{H_{i}\}$$

so $$\Im\{H_{i}\}-\Im\{H_{-i}\}=2\Im\{H_{i}\}=(\pi\coth(\pi)-1)$$

or

$$\Im\{H_{i}\}=\frac{\pi\coth(\pi)-1}{2}$$

or we know from Dr. Wolfgang Hintze's solution that $g=\Im\{H_{i}\}$


It would be interesting if we can find

$$H_{i}+H_{-i}\tag2$$

because from $(1)$ and $(2)$ we can find $H_{i}$ and $H_{-i}$.

Ali Shadhar
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