Evaluations of the higher order Polygamma functions

[DreamWeaver]

In this brief tutorial we evaluate the Trigamma, Tetragamma, and other higher order Polygamma functions at small rational arguments:$$(01) \quad \psi_{m \ge 1}(z) = (-1)^{m+1}m!\, \sum_{k=0}^{\infty} \frac{1}{(k+z)^{m+1}}$$We will have frequent need of the reflection formula, which is obtained by repeated differentiation of the reflection formula for the Digamma function:$$(02) \quad \psi_0(z) - \psi_0(1-z) = -\pi\cot \pi z \, \Rightarrow$$$$(03) \quad \psi_{m \ge 1}(z) + (-1)^{m+1}\psi_{m \ge 1}(1-z) = -\frac{d^m}{dz^m}\, \pi\cot \pi z$$From the definition (01) above, we have the trivial case:$$(04) \quad \psi_{m\ge 1}(1) = (-1)^{m+1}m!\, \zeta(m+1)$$. Hence$$\psi_1(1) = \zeta(2)$$$$\psi_2(1) = -2\, \zeta(3)$$$$\psi_3(1) = 6\, \zeta(4)$$$$\psi_4(1) = -24\, \zeta(5)$$$$\psi_5(1) = 120\, \zeta(6)$$$$\psi_6(1) = -720\, \zeta(7)$$
The case for $$z=1/2$$ is equally straightforward. Setting $$z=1/2$$ in (01) above we get:$$\psi_{m \ge 1} \left(\tfrac{1}{2}\right) = (-1)^{m+1}m!\, \sum_{k=0}^{\infty} \frac{1}{(k+1/2)^{m+1}}=
(-1)^{m+1}2^{m+1}\,m!\, \sum_{k=0}^{\infty} \frac{1}{(2k+1)^{m+1}} \equiv$$$$(-1)^{m+1}2^{m+1}\,m!\, \left[ \sum_{k=1}^{\infty} \frac{1}{k^{m+1}} - \frac{1}{2^{m+1}}\, \sum_{k=1}^{\infty} \frac{1}{k^{m+1}} \right]=$$$$(-1)^{m+1}2^{m+1}\,m!\, \left[ \left(1-\frac{1}{2^{m+1}} \right) \, \zeta(m+1) \right]$$
Hence$$(05) \quad \psi_{m\ge 1}\left(\tfrac{1}{2}\right) = (-1)^{m+1}(2^{m+1}-1)\,m!\, \zeta(m+1)$$. and$$\psi_1\left(\tfrac{1}{2}\right) = 3\zeta(2)$$$$\psi_2\left(\tfrac{1}{2}\right) = -14\, \zeta(3)$$$$\psi_3\left(\tfrac{1}{2}\right) = 90\, \zeta(4)$$$$\psi_4\left(\tfrac{1}{2}\right) = -744\, \zeta(5)$$$$\psi_5\left(\tfrac{1}{2}\right) = 7560\, \zeta(6)$$$$\psi_6\left(\tfrac{1}{2}\right) = -91440\, \zeta(7)$$

Comments and/or questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-evaluations-higher-order-polygamma-functions-8174.html

 

[DreamWeaver]

For a number of rational arguments whose denominator is $$\ge 3$$, we will need the duplication formula. This is obtained directly from the Legendre Duplication Formula for the Gamma function:$$(06) \quad \Gamma(2z)=\frac{2^{2z-1}}{\sqrt{\pi}}\,\Gamma(z)\,\Gamma\left( \tfrac{1}{2}+z\right)$$
Taking the logarithm of both sides, and then differentiating, we obtain the duplication formula for the Digamma function:$$\frac{d}{dz} \log\Gamma(2z) = 2\, \psi_0(2z) =$$ $$\frac{d}{dz} \Bigg\{ -\frac{1}{2}\log \pi+ (2z-1)\log 2 + \log \Gamma(z) +\log\Gamma\left( \tfrac{1}{2}+z\right) \Bigg\}=$$$$2\log 2 + \psi_0(z)+\psi_0\left( \tfrac{1}{2}+z\right)$$
Next, with the understanding that $$m\ge 1$$, we differentiate both sides m-times to obtain:
$$(07) \quad 2^{m+1}\psi_{m\ge 1}(2z)= \psi_{m\ge 1}(z)+\psi_{m\ge 1}\left( \tfrac{1}{2}+z\right)$$
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Setting $$z=1/4$$ in the duplication formula above gives:$$2^{m+1}\psi_{m\ge 1} \left( \tfrac{1}{2} \right) = \psi_{m\ge 1}\left( \tfrac{1}{4} \right)+
\psi_{m\ge 1}\left( \tfrac{3}{4} \right)$$Having already found a neat closed form for the 'half-argument' ($$z=1/2$$), we can write this as:$$\psi_{m\ge 1}\left( \tfrac{1}{4} \right)+
\psi_{m\ge 1}\left( \tfrac{3}{4} \right)= (-1)^{m+1}2^{m+1}(2^{m+1}-1)\, m!\, \zeta(m+1)$$Next, we consider the difference of Polygammas:$$\psi_{m\ge 1}\left( \tfrac{1}{4} \right)-
\psi_{m\ge 1}\left( \tfrac{3}{4} \right)= (-1)^{m+1}m!\, \Bigg\{ \sum_{k=0}^{\infty} \frac{1}{(k+1/4)^{m+1}} -
\sum_{k=0}^{\infty} \frac{1}{(k+3/4)^{m+1}}
\Bigg\}=$$$$(-1)^{m+1}2^{2m+2}\, m!\, \Bigg\{
\sum_{k=0}^{\infty} \frac{1}{(4k+1)^{m+1}} -
\sum_{k=0}^{\infty} \frac{1}{(4k+3)^{m+1}}
\Bigg\}=$$$$(-1)^{m+1}2^{2m+2}\, m!\,
\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^{m+1}}$$That last (alternating) series is a special case of the Dirichlet Beta function:$$\beta(x) = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x}$$Thus$$\psi_{m\ge 1}\left( \tfrac{1}{4} \right)+
\psi_{m\ge 1}\left( \tfrac{3}{4} \right)= (-1)^{m+1}2^{m+1}(2^{m+1}-1)\, m!\, \zeta(m+1)$$and$$\psi_{m\ge 1}\left( \tfrac{1}{4} \right)-
\psi_{m\ge 1}\left( \tfrac{3}{4} \right)= (-1)^{m+1}2^{2m+2}\, m!\, \beta(m+1)$$
Adding, respectively subtracting, those last two expressions gives
$$\psi_{m\ge 1}\left( \tfrac{1}{4} \right) = (-1)^{m+1}2^{m}\,m!\, \Bigg[ (2^{m+1}-1)\, \zeta(m+1) + 2^{m+1}\, \beta(m+1) \Bigg]$$
$$\psi_{m\ge 1}\left( \tfrac{3}{4} \right) = (-1)^{m+1}2^{m}\,m!\, \Bigg[ (2^{m+1}-1)\, \zeta(m+1) - 2^{m+1}\, \beta(m+1) \Bigg]$$
Thus$$\psi_{1}\left( \tfrac{1}{4} \right) = 6\,\zeta(2) + 8\,\beta(2) = \pi^2 + 8 G$$$$\psi_{1}\left( \tfrac{3}{4} \right) = 6\,\zeta(2) - 8\,\beta(2) = \pi^2 - 8 G$$$$\psi_{2}\left( \tfrac{1}{4} \right) = -56\,\zeta(3) - 64\,\beta(3) $$$$\psi_{2}\left( \tfrac{3}{4} \right) = -56\,\zeta(3) + 64\,\beta(3) $$$$\psi_{3}\left( \tfrac{1}{4} \right) = 720\,\zeta(4) + 768\,\beta(4) = \frac{\pi^4}{8
} + 768\, \beta(4)$$$$\psi_{3}\left( \tfrac{3}{4} \right) = 720\,\zeta(4) - 768\,\beta(4) = \frac{\pi^4}{8
} - 768\, \beta(4)$$$$\psi_{4}\left( \tfrac{1}{4} \right) = -11904\,\zeta(5) - 12288\,\beta(5)$$$$\psi_{4}\left( \tfrac{3}{4} \right) = -11904\,\zeta(5) + 12288\,\beta(5)$$$$\psi_{5}\left( \tfrac{1}{4} \right) = 241920\,\zeta(6) + 245760\,\beta(6) = \frac{\pi^6}{256}+ 245760\,\beta(6)$$$$\psi_{5}\left( \tfrac{3}{4} \right) = 241920\,\zeta(6) - 245760\,\beta(6) = \frac{\pi^6}{256}- 245760\,\beta(6)$$$$\psi_{6}\left( \tfrac{1}{4} \right) = -5852160\,\zeta(7) - 5898240\,\beta(7) $$$$\psi_{6}\left( \tfrac{3}{4} \right) = -5852160\,\zeta(7) + 5898240\,\beta(7) $$
Where I've used the classic Riemann Zeta values:$$\zeta(2) = \frac{\pi^2}{6}$$$$\zeta(4) = \frac{\pi^4}{90}$$$$\zeta(6) = \frac{\pi^6}{945}$$

 

[DreamWeaver]

As an aside, it's worth noting that the Dirichlet Beta functions of odd integer argument - $$\beta(1), \, \beta(3), \, \beta(5), \, \beta(7),$$ etc - can be evaluated explicitly by appealing to properties of the Polygamma function. On the one hand, we have$$\psi_{m\ge 1}\left( \tfrac{1}{4} \right)-
\psi_{m\ge 1}\left( \tfrac{3}{4} \right)= (-1)^{m+1}2^{2m+2}\, m!\, \beta(m+1) \, \Rightarrow$$$$\psi_{2m}\left( \tfrac{1}{4} \right)-
\psi_{2m}\left( \tfrac{3}{4} \right)= -2^{4m+2}\, (2m)!\, \beta(2m+1) $$Whereas on the other, the reflection formula - for $$m \ge 1$$ - gives:$$\psi_{2m}(z)-\psi_{2m}(1-z) = - \frac{d^{2m}}{dz^{2m}}\, \pi\cot \pi z$$Setting $$z=1/4$$ in that last relation, and equating it with the previous one gives:$$\beta(2m+1) = \frac{1}{2^{4m+2}\,(2m)!} \, \frac{d^{2m}}{dz^{2m}}\, \pi\cot \pi z \, \Bigg|_{z=1/4}$$Although here we are primarily concerned with the higher order Polygamma functions, it's worth noting that all of the relations above hold for the Digamma function. Hence the first few Dirichlet Beta functions of odd order are$$\beta(1) = \frac{\pi}{4}$$$$\beta(3) = \frac{\pi^3}{32}$$$$\beta(5) = \frac{5\pi^5}{1536}$$$$\beta(7) = \frac{61\pi^7}{184320}$$

 

[DreamWeaver]

In order to evaluate Polygamma functions whose (rational) arguments have a denominator $$> 4$$, a little more sophistry is required. To start with, we need to split the Polygammas into two distinct groups; one group for oddly indexed Polygammas, and one for evenly indexed Polygammas. The reason for this is that they will - in part - be evaluated in terms of Clausen functions, a function that itself differes according to parity.

Again, with the understanding that $$m \in \mathbb{Z} \ge 1$$:$$\text{Cl}_{2m}(\theta) = \sum_{k=1}^{\infty}\frac{\sin k\theta}{k^{2m}}$$$$\text{Cl}_{2m+1}(\theta) = \sum_{k=1}^{\infty}\frac{\cos k\theta}{k^{2m+1}}$$

Here's a brief sketch of the evaluation process...

Taking $$z=1/3$$ as an example, we express the Clausen function $$\text{Cl}_n(\pi z) = \text{Cl}_n(\pi/3)$$ as a sum/difference of $$6$$ partial sums, each of which is expressible as an $$(n-1)$$ order Polygamma function. This gives us one relation connecting the values of $$\psi_{n-1}\left( \tfrac{1}{6} \right), \, \psi_{n-1}\left( \tfrac{1}{3} \right), \, \psi_{n-1}\left( \tfrac{2}{3} \right), \,$$ and $$\psi_{n-1}\left( \tfrac{5}{6} \right)$$.

To get a second relation, we express the Riemann Zeta function of order $$n$$ as a sum of Polygamma functions of order $$(n-1)$$:$$\zeta(n) = \sum_{k=1}^{\infty} \frac{1}{k^n} = \sum_{j=1}^6 \sum_{k=0}^{\infty} \frac{1}{(6k+j)^n}$$To obtain a third relation connecting the very same arguments, we set $$z=1/3$$ in the Legendre-type duplication formula for the Polygamma function. If required, we also apply the reflection formula. The end result will be a direct evaluation for $$\psi_{n-1}\left( \tfrac{1}{6} \right), \, \psi_{n-1}\left( \tfrac{1}{3} \right), \, \psi_{n-1}\left( \tfrac{2}{3} \right), \,$$ and $$\psi_{n-1}\left( \tfrac{5}{6} \right)$$.

In cases where the denominator of the argument is slightly larger, say $$z=1/8$$, or $$z=1/12$$, we might require yet another (different!) relation linking related arguments of the Polygamma. One way around this problem is to set up a second correlation between Polygammas and a Clausen function of a different argument. In the 8-case, for example, we might consider$$\text{Cl}_n\left( \frac{\pi}{8} \right) = \sum_{j=1}^7 a_j\, \psi_{n-1}\left( \frac{j}{8} \right)$$and$$\text{Cl}_n\left( \frac{3\pi}{8} \right) = \sum_{j=1}^7 b_j\, \psi_{n-1}\left( \frac{j}{8} \right)$$In even higher order cases, we might set up a third Clausen function relation, although obviously we will endeavour to express Polygammas in as few Clausen functions as possible.That said, back to an evaluation... brb (Heidy)

 

[DreamWeaver]

Let $$m \ge 1$$, then we consider the sum$$\text{Cl}_{2m}\left( \frac{\pi}{3} \right) = \sum_{k=1}^{\infty}\frac{ \sin (\pi k/3) }{k^{2m}}$$By considering the first six terms, second six terms, third six terms, etc., we note that the first term in each sextet has coefficient $$\sin(\pi/3)$$, the second, $$\sin(2\pi/3)$$, etc. Indeed, each sextet is characterised by the following coefficients:$$\sin(\pi/3);\, \sin(2\pi/3); \, \sin(3\pi/3);\, \sin(4\pi/3); \, \sin(5\pi/3);\, \sin(6\pi/3) = $$$$\sin(\pi/3);\, \sin(2\pi/3); \, \sin(\pi);\, \sin(4\pi/3); \, \sin(5\pi/3);\, \sin(2\pi) \, \equiv$$$$\sin(\pi/3);\, \sin(2\pi/3); \, \sin(\pi);\, -\sin(2\pi/3); \, -\sin(\pi/3);\, \sin(2\pi) \, \equiv$$$$\sin(\pi/3);\, \sin(2\pi/3); \, 0;\, -\sin(2\pi/3); \, -\sin(\pi/3);\, 0 \, \equiv$$$$\sin(\pi/3);\, \sin(\pi/3); \, 0;\, -\sin(\pi/3); \, -\sin(\pi/3);\, 0 \, \equiv$$
Thus$$\text{Cl}_{2m}\left( \frac{\pi}{3} \right) = $$$$\sin \left(\frac{\pi}{3}\right) \sum_{k=0}^{\infty}\frac{1 }{(6k+1)^{2m}} +
\sin \left(\frac{\pi}{3}\right) \sum_{k=0}^{\infty}\frac{1 }{(6k+2)^{2m}} -$$$$\sin \left(\frac{\pi}{3}\right) \sum_{k=0}^{\infty}\frac{1 }{(6k+4)^{2m}} -
\sin \left(\frac{\pi}{3}\right) \sum_{k=0}^{\infty}\frac{1 }{(6k+5)^{2m}} = $$
$$\frac{\sqrt{3}}{2\, (6^{2m})}\,\sum_{k=0}^{\infty} \Bigg\{
\frac{1 }{(k+1/6)^{2m}}+
\frac{1 }{(k+1/3)^{2m}}-
\frac{1 }{(k+2/3)^{2m}}-
\frac{1 }{(k+5/6)^{2m}}
\Bigg\}=$$$$\frac{\sqrt{3}}{2\, (6^{2m})}\, \frac{(-1)^{2m}}{(2m-1)!}\, \Bigg\{
\psi_{2m-1}\left(\frac{1}{6}\right) +
\psi_{2m-1}\left(\frac{1}{3}\right) -
\psi_{2m-1}\left(\frac{2}{3}\right) -
\psi_{2m-1}\left(\frac{5}{6}\right)
\Bigg\}
$$So our first relation is:
$$(08) \quad \psi_{2m-1}\left(\frac{1}{6}\right) +
\psi_{2m-1}\left(\frac{1}{3}\right) -
\psi_{2m-1}\left(\frac{2}{3}\right) -
\psi_{2m-1}\left(\frac{5}{6}\right) = $$$$\frac{2\, (6^{2m})\, (2m-1)!}{ \sqrt{3} } \, \text{Cl}_{2m}\left( \frac{\pi}{3} \right)$$

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For the second relation, we write:$$\zeta(2m) = \sum_{k=1}^{\infty}\frac{1}{k^{2m}} \equiv \sum_{j=1}^6\, \sum_{k=0}^{\infty}\frac{1}{(6k+j)^{2m}} =$$$$\frac{1}{6^{2m}}\, \sum_{j=1}^6\, \sum_{k=0}^{\infty}\frac{1}{(k+j/6)^{2m}} =$$$$\frac{1}{6^{2m}\, (2m-1)! }\, \sum_{j=1}^6\, \psi_{2m-1}\left(\frac{j}{6}\right) \, \Rightarrow$$$$6^{2m}\, (2m-1)!\, \zeta(2m) = \sum_{j=1}^6\, \psi_{2m-1}\left(\frac{j}{6}\right) = $$$$\psi_{2m-1}\left(\frac{1}{2}\right) + \psi_{2m-1}\left(1\right) + \sum_{j=1, 2, 4, 5}\, \psi_{2m-1}\left(\frac{j}{6}\right)=$$$$(2m-1)! \, \zeta(2m) + (2^{2m} - 1)\,(2m-1)! \, \zeta(2m) +
\sum_{j=1, 2, 4, 5}\, \psi_{2m-1}\left(\frac{j}{6}\right)=$$$$2^{2m}\, (2m-1)! \, \zeta(2m) +
\sum_{j=1, 2, 4, 5}\, \psi_{2m-1}\left(\frac{j}{6}\right)$$Hence$$\sum_{j=1, 2, 4, 5}\, \psi_{2m-1}\left(\frac{j}{6}\right) = (6^{2m}-2^{2m})\, (2m-1)!\, \zeta(2m)$$Or equivalently$$(09) \quad \psi_{2m-1}\left(\frac{1}{6}\right) +
\psi_{2m-1}\left(\frac{1}{3}\right) +
\psi_{2m-1}\left(\frac{2}{3}\right) +
\psi_{2m-1}\left(\frac{5}{6}\right) =$$$$ 2^{2m}(3^{2m}-1)\, (2m-1)!\, \zeta(2m)$$

---------------------------------------
Adding and subtracting the Polygamma relations (08) and (09) gives:$$(10) \quad \psi_{2m-1}\left(\frac{1}{6}\right) +
\psi_{2m-1}\left(\frac{1}{3}\right) = $$$$ 2^{2m-1}(3^{2m}-1)\, (2m-1)!\, \zeta(2m) +
\frac{(6^{2m})\, (2m-1)!}{ \sqrt{3} } \, \text{Cl}_{2m}\left( \frac{\pi}{3} \right)
$$and$$(11) \quad \psi_{2m-1}\left(\frac{2}{3}\right) +
\psi_{2m-1}\left(\frac{5}{6}\right) = $$$$ 2^{2m-1}(3^{2m}-1)\, (2m-1)!\, \zeta(2m) -
\frac{(6^{2m})\, (2m-1)!}{ \sqrt{3} } \, \text{Cl}_{2m}\left( \frac{\pi}{3} \right)
$$

Almost done... (Heidy)

 

[DreamWeaver]

Setting $$z=1/6$$ in the Legendre-type duplication formula - (07) above - we get:$$2^{2m} \, \psi_{2m-1}\left(\frac{1}{3}\right) = \psi_{2m-1}\left(\frac{1}{6}\right) + \psi_{2m-1}\left(\frac{2}{3}\right)$$Rearranging the terms to...$$\psi_{2m-1}\left(\frac{1}{6}\right) = \psi_{2m-1}\left(\frac{2}{3}\right) - 2^{2m} \, \psi_{2m-1}\left(\frac{1}{3}\right) $$...and then substituting this back into (10) gives:$$(12) \quad
\psi_{2m-1}\left(\frac{2}{3}\right) + (1-2^{2m}) \, \psi_{2m-1}\left(\frac{1}{3}\right) =$$$$ 2^{2m-1}\, (3^{2m}-1)\, (2m-1)!\, \zeta(2m) + \frac{6^{2m}(2m-1)!}{\sqrt{3}}\, \text{Cl}_{2m}\left( \frac{\pi}{3} \right)$$
This is exactly what we were looking for all along; an expression that can be reduced to a single Polygamma function, by use of the reflection formula...$$(13) \quad \psi_{2m-1}\left( \frac{1}{3} \right) +
\psi_{2m-1}\left( \frac{2}{3} \right) = - \frac{d^{2m-1}}{dz^{2m-1}}\, \pi\cot \pi z \, \Bigg|_{z=1/3}$$Subtracting $$(13)$$ from $$(12)$$ gives:$$\psi_{2m-1}\left( \frac{1}{3} \right) = $$$$- \frac{1}{2^{2m}}\, \Bigg\{
2^{2m-1}\, (3^{2m}-1)\, (2m-1)!\, \zeta(2m) + \frac{6^{2m}(2m-1)!}{\sqrt{3}}\, \text{Cl}_{2m}\left( \frac{\pi}{3} \right)
+
\frac{d^{2m-1}}{dz^{2m-1}}\, \pi\cot \pi z \, \Bigg|_{z=1/3}
\Bigg\}=$$$$-\frac{3^{2m}\, (2m-1)!}{2} \Bigg\{ (1-3^{-2m})\, \zeta(2m) + \frac{2}{\sqrt{3}}\,\text{Cl}_{2m}\left( \frac{\pi}{3}\right) \,
\Bigg\} -
\frac{1}{2^{2m}}\,\frac{d^{2m-1}}{dz^{2m-1}}\, \pi\cot \pi z \, \Bigg|_{z=1/3}$$---------------------------------------

$$(14) \quad \psi_{2m-1}\left(\frac{1}{3}\right) = -\frac{1}{2^{2m}}\,\frac{d^{2m-1}}{dz^{2m-1}}\, \pi\cot \pi z \, \Bigg|_{z=1/3}$$$$-\frac{3^{2m}\, (2m-1)!}{2} \Bigg\{ (1-3^{-2m})\, \zeta(2m) + \frac{2}{\sqrt{3}}\,\text{Cl}_{2m}\left( \frac{\pi}{3}\right) \,
\Bigg\}
$$

$$(15) \quad \psi_{2m-1}\left(\frac{2}{3}\right) = \frac{(1-2^{2m})}{2^{2m}}\,\frac{d^{2m-1}}{dz^{2m-1}}\, \pi\cot \pi z \, \Bigg|_{z=1/3}$$$$+ \frac{3^{2m}\, (2m-1)!}{2} \Bigg\{ (1-3^{-2m})\, \zeta(2m) + \frac{2}{\sqrt{3}}\,\text{Cl}_{2m}\left( \frac{\pi}{3}\right) \,
\Bigg\}
$$

More soon... ish. (Heidy)