Dtyjlui

Is there an accepted analytic continuation of $sum_{n=1}^m frac{1}{n}$? Even a continuation to positive reals would be of interested, though negative and complex arguments would also be interesting.

I don't have a specific application in mind, but I'd very much like to understand how / if such a continuation could be accomplished. I've Googled but haven't come up with anything meaningful - perhaps because it's not possible?

ADDENDUM

@Noble below suggests $frac{Gamma'(x)}{Gamma(x)}$. But this produces the following mismatched plots:

Can anyone explain?

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

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.

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.

2. 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

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

3. 
   $H_{z}$ 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?