Krdytkyu

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

$$zetaleft(2,frac14right)~=~pi^2+8Gtag1$$

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

$$begin{align*}
zeta(2)=sum_{n=1}^infty frac1{n^2}&=frac1{1^2}+frac1{2^2}+frac1{3^2}+sum_{n=1}^inftyfrac1{(4n)^2}+frac1{(4n+1)^2}+frac1{(4n+2)^2}+frac1{(4n+3)^2}\
&=frac1{16}zeta(2)+sum_{n=0}^infty frac1{4(2n+1)^2}+frac1{(4n+1)^2}+frac1{(4n+3)^2}\
&=frac1{16}zeta(2)+frac14lambda(2)+sum_{n=0}^infty frac1{(4n+1)^2}+frac12left[frac1{(2n+1)^2}-frac{(-1)^n}{(2n+1)^2}right]\
&=frac1{16}zeta(2)+frac14lambda(2)+frac12[lambda(2)-beta(2)]+sum_{n=0}^infty frac1{(4n+1)^2}\
therefore~sum_{n=0}^infty frac1{(4n+1)^2}&=frac{15}{16}zeta(2)-frac34lambda(2)+frac12beta(2)\
&=frac{15}{16}zeta(2)-frac34frac34zeta(2)+frac12 G\&=frac1{16}pi^2+frac12 G
end{align*}$$

$$therefore~zetaleft(2,frac14right)~=~sum_{n=0}^inftyfrac1{left(n+frac14right)^2}~=~pi^2+8G$$

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

$$begin{align*}
psi^{(n)(z)}&=(-1)^{n+1}n!sum_{k=0}^inftyfrac1{(z+k)^{n+1}}\
&=(-1)^{n+1}n!zeta(n+1,z)
end{align*}$$

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

$$(-1)^{1+1}(1!)psi^{(1)}left(frac14right)=zetaleft(1+1,frac14right)Rightarrow psi^{(1)}left(frac14right)=zetaleft(2,frac14right)$$

We can use the following representation of the digamma function from here, namely:
$$psi(s+1)=-gamma+int_0^1 frac{1-x^s}{1-x}dxRightarrow psi_1(s+1)=int_0^1 frac{x^{s}ln x}{x-1}dx$$
Above follows since $frac{d}{dz}psi(z)=psi_1{(z)} $, so we can rewrite our desired value as:
$$psi_1left(frac14right)=int_0^1 frac{x^{1/4-1} ln x}{x-1}dx$$
Notice that $frac{d}{dx}left(4x^{frac{1}{4}}right)=x^{frac14-1}$, so we can easily substitute $x^{frac14}=tRightarrow x=t^4$ to get: $$psi_1left(frac14right)=16int_0^1 frac{ln t}{t^4-1}dt$$
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.

$$frac{1}{x^4-1}=frac12 frac{(x^2+1)-(x^2-1)}{(x^2+1)(x^2-1)}=frac12 left(frac1{x^2-1}-frac{1}{x^2+1}right)$$
$$Rightarrow psi_1left(frac14right)=8int_0^1frac{ln t}{t^2-1}dt -8int_0^1 frac{ln t}{t^2+1}dt=boxed{pi^2 +8G}$$
I am pretty sure you can prove that the first integral equals to $frac{pi^2}{8}$, but here are found many proofs.

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

$$zeta(s,q)~=~frac1{Gamma(s)}int_0^infty frac{t^{s-1}e^{-qt}}{1-e^{-t}}mathrm dttag1$$

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

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

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.