Analytic continuation of harmonic series

Let’s try it in an elementary manner

  1. We can use the defining recursion of the harmonic number valid for $nin Z^{+}$

$$H_{n} = H_{n-1} + frac{1}{n}, H_{1}=1tag{1a}$$

also for any complex $z$

$$H_{z} = H_{z-1} + frac{1}{z}, H_{1}=1tag{1b}$$

For instance for $z=1$ we obtain $$H_{1} = H_{0} + frac{1}{1}$$

from which we conclude that $H_{0}=0$.

If we try to find $H_{-1}$ we encounter the problem that from $H_0 = 0 = lim_{zto0}(H_{-1+z} + frac{1}{z})$ we find that $H_{z} simeq frac{1}{z}$ for $zsimeq 0$. In other words, $H_{z}$ has a simple pole at $z=-1$.

Hence we cannot continue in this manner to go to further into the region of negative $z$, so let us move to the following general approach.

  1. Starting with this formula for the harmonic number which is valid for $nin Z^{+}$

$$H_{n} = frac{1}{2}+ … + frac{1}{n}\\=frac{1}{1}+ frac{1}{2}+ … + frac{1}{n} +frac{1}{1+n}+frac{1}{n+2} + … \;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;-frac{1}{1+n}- frac{1}{n+2} + …\=sum_{k=0}^infty left(frac{1}{k}-frac{1}{ (k+n)}right)tag{2}$$

The sum can be written as

$$H_{n}= sum_{k=1}^infty frac{n}{k (k+n)}tag{3}$$

and this can be extended immediately to complex values $z$ in place of $n$

$$H_{z}= sum_{k=1}^infty frac{z}{k (k+z)}=sum_{k=1}^infty left(frac{1}{k} -frac{1}{k+z}right)tag{4}$$

This sum is convergent (the proof is left to the reader) for any $z$ except for $z=-1, -2, …$ where $H_{z}$ has simple poles with residue $-1$.

Hence $(4)$ gives the analytic continuation.

For instance close to $z=0$ we have as in 1. that

$$H_{z} simeq z sum_{k=1}^infty frac{1}{k^2} = z;zeta(2) =z;frac{pi^2}{6}to 0 $$

We can also derive an integral representation from the second form of $(4)$ writing

$$frac{1}{k} -frac{1}{k+z} =int_0^1 (x^{k-1}-x^{z+k-1}),dx $$

Performing the sum under the integral is just doing a geometric sum and gives

$$H_{z} = int_0^1 frac{1-x^{z}}{1-x},dx tag{5}$$

  1. $H_{z}$ at negativ half integers ($z = -frac{1}{2}, -frac{3}{2}, …$)

These can be calculated from $(1b)$ as soon as $H_{frac{1}{2}}$ is known.

So let us calculate $H_frac{1}{2}$.

Consider

$$H_{2n} = frac{1}{1} + frac{1}{2} + frac{1}{3} + … + frac{1}{2n}$$

Splitting even and odd terms gives

$$H_{2n}=
frac{1}{1} + frac{1}{3} + frac{1}{5} + … + frac{1}{2n-1}\+
frac{1}{2} + frac{1}{4} + … + frac{1}{2n}\=
sum_{k=1}^n frac{1}{2k-1} + frac{1}{2} H_{n}tag{6}$$

Now for the sum of the odd terms we write as in $(1)$

$$O_{n} = sum_{k=1}^infty left(frac{1}{2k-1} – frac{1}{2(n+k)-1}right)tag{7}$$

This can be anlytically continued to any complex $nto z$.

Replacing as before the summand by an integral and doing the summation under the integral gives

$$O_{z} = int_0^1 frac{1-x^{2z}}{1-x^2},dxtag{8} $$

Substituting $x to sqrt{t}$ we find

$$O_{z} = frac{1}{2}int_0^1 frac{1}{sqrt{t}}frac{1-t^{z}}{1-t},dt\=
frac{1}{2}int_0^1 frac{1-t^{z-frac{1}{2}}}{1-t},dt-
frac{1}{2}int_0^1 frac{1-t^{-frac{1}{2}}}{1-t},dt\
=frac{1}{2}H_{z-frac{1}{2}}+log{2}tag{9}$$

Hence $(6)$ can be written as

$$H_{2z} = frac{1}{2} H_{z} +frac{1}{2} H_{z-frac{1}{2}}+log{2}tag{10} $$

Letting $z=1$ this gives

$$H_{2} = frac{1}{2} H_{1} +frac{1}{2} H_{frac{1}{2}}+log{2} $$

from which we deduce finally

$$H_{frac{1}{2}} = 2(1-log{2})simeq 0.613706 tag{11}$$

EDIT

Altenatively, the calculation of $H_{frac{1}{2}}$ can be done using $(5)$ with the substitution $(xto t^2)$:

$$H_{frac{1}{2}} = int_0^1 frac{1-x^{frac{1}{2}}}{1-x},dx
= 2int_0^1 t frac{(1-t)}{{1-t^2}},dt = 2int_0^1 t frac{(1-t)}{(1+t)(1-t)},dt \=2int_0^1 frac{t}{{1+t}},dt=2int_0^1 frac{1+t}{{1+t}},dt -2int_0^1 frac{1}{{1+t}},dt = 2 – 2 log(2)$$

and we have recovered $(11)$.

As an exercise calculate $H_{frac{1}{n}}$ for $n =3, 4,…$.

I found that Mathematica returns explicit expression up to $n=12$ except for the case $n=5$. I have not yet unerstood the reason for this exception. Maybe someone else can explain it?

I am not sure if this is what you meant, but Wolfram Alpha has an analytic formula for the $n^{text{th}}$ harmonic number:

enter image description here

Here, the digamma function is $psi_0(x)=frac{Gamma'(x)}{Gamma(x)}$, which I believe is defined for all numbers in the complex plane except for negative real integers.

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