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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 $(1)$?

Thanks in advance!

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

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

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    $\begingroup$ Isn't this $\psi_1 \left(\frac14\right)$ (trigamma function)? See here: mathworld.wolfram.com/TrigammaFunction.html $\endgroup$ – カカロット Jan 5 at 19:24
  • $\begingroup$ @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. $\endgroup$ – mrtaurho Jan 5 at 19:28
  • $\begingroup$ I see, so basically the question is to prove that $\psi\left(\frac14\right) =\pi^2 +8G$, right? $\endgroup$ – カカロット Jan 5 at 19:29
  • $\begingroup$ @Zacky Yes and no. Basically it is the same question, yes even though I am more interested in Zeta Functions at the moment. $\endgroup$ – mrtaurho Jan 5 at 19:30
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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.

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    $\begingroup$ 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. $\endgroup$ – mrtaurho Jan 5 at 22:01
  • $\begingroup$ ^_^ 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. $\endgroup$ – カカロット Jan 5 at 22:03
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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.

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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.

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