# 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:

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.