# Alternative proof for $\zeta\left(2,\frac14\right)=\psi^{(1)}\left(\frac14\right)=\pi^2+8G$

On the German Wikipedia page of the Hurwitz Zeta Function I have come across the following formula

$$\zeta\left(2,\frac14\right)~=~\pi^2+8G\tag1$$

Where $$G$$ is Catalan's Constant. Even though I was able proving $$(1)$$ I am dissatisfied with my own attempt since it is heavily relying on several relatives of the Riemann Zeta Functions. However, first of all I will present my own solution. Starting with the series representation of $$\zeta(2)$$ we get

\begin{align*} \zeta(2)=\sum_{n=1}^\infty \frac1{n^2}&=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\sum_{n=1}^\infty\frac1{(4n)^2}+\frac1{(4n+1)^2}+\frac1{(4n+2)^2}+\frac1{(4n+3)^2}\\ &=\frac1{16}\zeta(2)+\sum_{n=0}^\infty \frac1{4(2n+1)^2}+\frac1{(4n+1)^2}+\frac1{(4n+3)^2}\\ &=\frac1{16}\zeta(2)+\frac14\lambda(2)+\sum_{n=0}^\infty \frac1{(4n+1)^2}+\frac12\left[\frac1{(2n+1)^2}-\frac{(-1)^n}{(2n+1)^2}\right]\\ &=\frac1{16}\zeta(2)+\frac14\lambda(2)+\frac12[\lambda(2)-\beta(2)]+\sum_{n=0}^\infty \frac1{(4n+1)^2}\\ \therefore~\sum_{n=0}^\infty \frac1{(4n+1)^2}&=\frac{15}{16}\zeta(2)-\frac34\lambda(2)+\frac12\beta(2)\\ &=\frac{15}{16}\zeta(2)-\frac34\frac34\zeta(2)+\frac12 G\\&=\frac1{16}\pi^2+\frac12 G \end{align*}

$$\therefore~\zeta\left(2,\frac14\right)~=~\sum_{n=0}^\infty\frac1{\left(n+\frac14\right)^2}~=~\pi^2+8G$$

Where $$\zeta(s)$$ dnotes the Riemann Zeta Function, $$\lambda(s)$$ the Dirichlet Lambda Function, $$\beta(s)$$ the Dirichlet Beta Function. Moreover the relation $$\lambda(s)=(1-2^{-s})\zeta(s)$$ and the well-known values $$\zeta(2)=\frac{\pi^2}6$$ and $$\beta(2)=G$$ were used.

I am suspicious about the heavy usage of the whole Zeta Function machinery and I am curious if there exists a shorter and more elegant way of proving $$(1)$$?

EDIT

As Zacky pointed out within the comments this special value of the Hurwitz Zeta Function can also be interpreted as a particular value of the Trigamma Function hence in general we got a the following series expansion of the Polygamma Function

\begin{align*} \psi^{(n)(z)}&=(-1)^{n+1}n!\sum_{k=0}^\infty\frac1{(z+k)^{n+1}}\\ &=(-1)^{n+1}n!\zeta(n+1,z) \end{align*}

So for $$n=1$$ and $$z=\frac14$$ it directly follows that

$$(-1)^{1+1}(1!)\psi^{(1)}\left(\frac14\right)=\zeta\left(1+1,\frac14\right)\Rightarrow \psi^{(1)}\left(\frac14\right)=\zeta\left(2,\frac14\right)$$

• Isn't this $\psi_1 \left(\frac14\right)$ (trigamma function)? See here: mathworld.wolfram.com/TrigammaFunction.html – カカロット Jan 5 at 19:24
• @Zacky This is due the more general fact that $$\psi^{(n)}(z)~=~(-1)^{n+1}n!\zeta(n+1,z)$$ which can be shown directly with the series representation of both functions. But thank I you, I have forgotten about this relation. – mrtaurho Jan 5 at 19:28
• I see, so basically the question is to prove that $\psi\left(\frac14\right) =\pi^2 +8G$, right? – カカロット Jan 5 at 19:29
• @Zacky Yes and no. Basically it is the same question, yes even though I am more interested in Zeta Functions at the moment. – mrtaurho Jan 5 at 19:30

We can use the following representation of the digamma function from here, namely: $$\psi(s+1)=-\gamma+\int_0^1 \frac{1-x^s}{1-x}dx\Rightarrow \psi_1(s+1)=\int_0^1 \frac{x^{s}\ln x}{x-1}dx$$ Above follows since $$\frac{d}{dz}\psi(z)=\psi_1{(z)}$$, so we can rewrite our desired value as: $$\psi_1\left(\frac14\right)=\int_0^1 \frac{x^{1/4-1} \ln x}{x-1}dx$$ Notice that $$\frac{d}{dx}\left(4x^{\frac{1}{4}}\right)=x^{\frac14-1}$$, so we can easily substitute $$x^{\frac14}=t\Rightarrow x=t^4$$ to get: $$\psi_1\left(\frac14\right)=16\int_0^1 \frac{\ln t}{t^4-1}dt$$ We might already recall that we saw before something similar, $$\int_0^1 \frac{\ln x}{1+x^2}dx=-G$$, thus let's try to get there.

$$\frac{1}{x^4-1}=\frac12 \frac{(x^2+1)-(x^2-1)}{(x^2+1)(x^2-1)}=\frac12 \left(\frac1{x^2-1}-\frac{1}{x^2+1}\right)$$ $$\Rightarrow \psi_1\left(\frac14\right)=8\int_0^1\frac{\ln t}{t^2-1}dt -8\int_0^1 \frac{\ln t}{t^2+1}dt=\boxed{\pi^2 +8G}$$ I am pretty sure you can prove that the first integral equals to $$\frac{\pi^2}{8}$$, but here are found many proofs.

• Well done (+1) This is the kind of answer I had in mind while formulating my question. Okay, to be honest I was not thinking about the Trigamma Function since I had forgotten about this relation but anyway thank you. – mrtaurho Jan 5 at 22:01
• ^_^ Now that I think of it, maybe it was more directly to use this: en.wikipedia.org/wiki/…, but I am more familiar with trigamma function than to Hurwitz's Zeta function. – カカロット Jan 5 at 22:03

Inspired by Zacky's comment on the integral representation of the Hurwitz Zeta Function I have found another pretty straightforward way. First we notice that

$$\zeta(s,q)~=~\frac1{\Gamma(s)}\int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}\mathrm dt\tag1$$

Now substitute $$s=2$$ and $$q=\frac14$$ followed by $$u=e^{-t}$$ to get

$$\mathfrak I=\int_0^{\infty}\frac{te^{-t/4}}{1-e^{-t}}\mathrm dt=\int_0^1\frac{u^{1/4-1}\cdot \log(u)}{u-1}\mathrm du$$

And now we are at the same point from where Zacky deduced the right value via well-known integrals. Hence his method was quite elegant I will not repeat his solution but just refer to it. Basically using $$(1)$$ does not force us to rely on the Trigamma Function but instead leaves us all along with the Hurwitz Zeta Function and some nice integrals.

We can use the generalization identity

$$\psi^{(n)}(a)=-\int_0^1\frac{x^{a-1}\ln^n(x)}{1-x}\ dx$$

set $$n=1$$ and $$a=1/4$$ we get

$$\psi^{(1)}\left(\frac14\right)=-\int_0^1\frac{x^{1/4-1}\ln(x)}{1-x}\ dx$$

which is the same integral @Zacky got.