polygamma at a quarter

the trick is to write the legendre duplication and euler reflection formulas $2^{2n+1} \psi_{2n} (2z)=\psi_{2n} (z)+\psi_{2n} (z+1/2)\\\psi_{2n} (1-z)-\psi_{2n}(z)=\pi^{2n+1} \dfrac{d^{2n}}{dx^{2n}}\cot(x)\large|_{x=\pi/4}$ putting n=2, we first calculate $\dfrac{d^{4}}{dx^{4}}\cot(x)=-\dfrac{d^{3}}{dx^{3}}\csc^2(x)=2\dfrac{d^{2}}{dx^{2}}\cot(x) \csc^2(x)=\\-2\dfrac{d}{dx}(2\csc^4(x)-\csc^2(x)+\cot^2(x)\csc^2(x))=2(-3 \cot(x) \csc^2(x) + \cot^3(x) \csc^2(x) + 11 \cot(x) \csc^4(x))$ evaluating at $x=\frac{\pi}{4}$ , (replace all $\cot(x)$ with 1 and $\csc^2(x)$ with 2) we get $80$ . so we have the equations $32 \psi_{4} (1/2)=\psi_{4} (1/4)+\psi_{4} (3/4)\to -32(4!)(2^5-1)\zeta(5)=\psi_{4} (1/4)+\psi_{4} (3/4)\\\psi_{4} (3/4)-\psi_{4} (1/4)=80\pi^5$ a result from a problem linked in the problem was used. we can subtract both equations and divide by 2 to get $\psi_{4} (1/4)=-2^3(1488 \zeta(5)+5\pi^5)$