# Trigamma identity $4\,\psi_1\!\left(\frac15\right)+\psi_1\!\left(\frac25\right)-\psi_1\!\left(\frac1{10}\right)=\frac{4\pi^2}{\phi\,\sqrt5}.$

I heuristically discovered the following identity for the trigamma function, that I could not find in any tables or papers or infer from existing formulae (e.g. [1], [2], [3], [4], [5], [6]): $$4\,\psi_1\!\left(\frac15\right)+\psi_1\!\left(\frac25\right)-\psi_1\!\left(\frac1{10}\right)=\frac{4\pi^2}{\phi\,\sqrt5}.\tag1$$ It also seems to be unknown to Mathematica, but numerically checks with at least $20000$ decimal digits. It might be provable through some application of reflection and multiplication theorems, but I couldn't do this.

Please suggest how to prove it.

Update: Another identity is $$3\,\psi_1\!\left(\frac1{12}\right)-30\,\psi_1\!\left(\frac13\right)=120\,G+\left(6\sqrt3-8\right)\pi^2.\tag2$$

• Vladi, when are you publishing a book with all your crazy special function identities? – nbubis Aug 15 '14 at 6:17
• Could you tell me (with details) what is your heuristics for finding so interesting identities ? Thanks. – Claude Leibovici Aug 15 '14 at 6:22
• @ClaudeLeibovici In this case I used Mathematica function FindIntegerNullVector. For a list of approximate real numbers it tries to find their linear combination with integers coefficients that is zero (with a certain precision). But it starts giving false positives if the list of numbers is long, so you need to carefully select plausible candidates that might partitipate in a conjectured linear identity. For multiplicative identities (like those I posted for hypergeometric function) you just work with logs instead. – Vladimir Reshetnikov Aug 15 '14 at 15:46
• @ClaudeLeibovici There are some other methods as well. You could ask at mathematica.stackexchange.com and let me know. Then I'll provide more details and code in an answer, these comments are too short for that. – Vladimir Reshetnikov Aug 15 '14 at 16:00
• @nbubis Not a book, but I have an idea to compile and publish the list of values of the polygamma function at rational points expressed through $\pi, G, \zeta(n)$ and values of the polygamma function of simpler arguments. – Vladimir Reshetnikov Aug 15 '14 at 23:49

I will use only the standard reflection and duplication identities: \begin{align} &\psi_1(z)+\psi_1(1-z)=\frac{\pi^2}{\sin^2\pi z},\\ &\psi_1(z)+\psi_1\bigl(z+\text{$\frac12$}\bigr)=4\psi_1(2z). \end{align} Use the first of them to replace $\psi_1\bigl(\frac1{10}\bigr)$ by $-\psi_1\bigl(\frac9{10}\bigr)+\displaystyle\frac{\pi^2}{\sin^2\frac{\pi}{10}}$. Then use the second to replace $\psi_1\bigl(\frac25\bigr)+\psi_1\bigl(\frac{9}{10}\bigr)$ by $4\psi_1\bigl(\frac45\bigr)$. Finally use the first the second time to replace $4\psi_1\bigl(\frac15\bigr)+4\psi_1\bigl(\frac45\bigr)$ by $\displaystyle\frac{4\pi^2}{\sin^2\frac{\pi}{5}}$. In this way we obtain $$4\psi_1\left(\frac15\right)+\psi_1\left(\frac25\right)-\psi_1\left(\frac1{10}\right)=\frac{4\pi^2}{\sin^2\frac{\pi}{5}}-\frac{\pi^2}{\sin^2\frac{\pi}{10}},$$ and the rest is straightforward.
As for the second identity, we will use in addition that $$\sum_{k=0}^2\psi_1\bigl(z+\text{\frac{k}3}\bigr)=9\psi_1(3z),\qquad \sum_{k=0}^3\psi_1\bigl(z+\text{\frac{k}4}\bigr)=16\psi_1(4z).$$ This allows to write $$\psi_1\left(\frac1{12}\right)+\psi_1\left(\frac5{12}\right)+\psi_1\left(\frac3{4}\right)=9\psi_1\left(\frac14\right),$$ $$\psi_1\left(\frac1{12}\right)+\psi_1\left(\frac13\right)+\psi_1\left(\frac7{12}\right)+\psi_1\left(\frac5{6}\right)=16\psi_1\left(\frac13\right).$$ Adding the two identities, we find $$2\psi_1\left(\frac1{12}\right)+\frac{\pi^2}{\sin^2\frac{5\pi}{12}}+4\psi_1\left(\frac23\right)+\psi_1\left(\frac34\right)=9\psi_1\left(\frac14\right)+16\psi_1\left(\frac13\right).$$ Now using two times the reflection formula, we can rewrite the last identity as $$2\psi_1\left(\frac1{12}\right)+\frac{\pi^2}{\sin^2\frac{5\pi}{12}}+\frac{4\pi^2}{\sin^2\frac{2\pi}{3}}+\frac{\pi^2}{\sin^2\frac{3\pi}{4}}=10\psi_1\left(\frac14\right)+20\psi_1\left(\frac13\right),$$ or, equivalently, $$2\psi_1\left(\frac1{12}\right)-20\psi_1\left(\frac13\right)=10\psi_1\left(\frac14\right)-\frac{46-12\sqrt3}{3}\pi^2.$$ Now it suffices to use the formula (4) from here.
• Thanks! I was able to prove the identity $(2)$ in the same way. – Vladimir Reshetnikov Aug 15 '14 at 21:48
Consider the multiplication formula for the polygamma function $$\psi_n(mz)=\frac{1}{m^{n+1}}\sum_{k=0}^{m-1}\psi_n\left(z+\frac km\right)\quad;\quad\text{for}\ n\ge1\tag1$$ and its reflection formula $$\psi_n(1-z)+(-1)^{n+1}\psi_n(1-z)=(-1)^n\pi\frac{d^n}{dz^n}\cot\pi z.\tag2$$ Using $(1)$ by setting $n=1,\ m=2,$ and $z=\dfrac1{10}$, we obtain \begin{align} \psi_1\left(\frac15\right)&=\frac{1}{4}\left[\psi_1\left(\frac1{10}\right)+\psi_1\left(\frac35\right)\right]\\ \psi_1\left(\frac35\right)&=4\psi_1\left(\frac15\right)-\psi_1\left(\frac1{10}\right)\tag3 \end{align} then using $(3)$ and with helping $(2)$, we obtain \begin{align} 4\psi_1\left(\frac15\right)+\psi_1\left(\frac25\right)-\psi_1\left(\frac1{10}\right)&=\psi_1\left(\frac35\right)+\psi_1\left(\frac25\right)\\ &=-\pi\left.\frac{d}{dz}\cot\pi z\right|_{z=\frac25}\\ &=\frac{8\pi^2}{5+\sqrt{5}}\\ &=\frac{4\pi^2}{\phi\sqrt{5}}.\tag{Q.E.D.} \end{align}