Polygamma at half

For $n\in \mathbb{Z^+}$ ,

$\displaystyle \begin{aligned} \int_0^1 \dfrac{\ln^{n}x}{(x-1)\sqrt{x}}\; dx &= 2^{n+1} \int_0^1 \dfrac{\ln^n x}{(x^2-1)}\; dx \\ &= -2^n \left[ \int_0^1 \dfrac{\ln^n x}{1-x}\; dx+\int_0^1 \dfrac{\ln^ n x}{1+x}\; dx \right] \\ &= -2^n \left[ \sum_{m=0}^\infty \int_0^1 x^m \ln^n x\; dx +\sum_{m=0}^\infty (-1)^m \int_0^1 x^m \ln^n x \; dx\right] \\ &= -2^n \left[ \sum_{m=0}^\infty \dfrac{n! (-1)^n}{(m+1)^{n+1}} +\sum_{m=0}^\infty \dfrac{n! (-1)^{m+n}}{(m+1)^{n+1}}\right] \\ &= 2^n (-1)^{n+1}\Gamma(n+1)\zeta(n+1)(2-2^{-n}) \\ &= (-1)^{n+1}(2^{n+1}-1)\Gamma(n+1)\zeta(n+1) \\ &= (-1)^{n+1}(2^{n+1}-1)\int_0^\infty \dfrac{x^n}{e^x-1}\; dx \end{aligned}$

This makes the answer $\boxed{1729+1730=3459}$ after substituting for $n=1729$

consider the integral representations of $\psi_n(z)=\int_0^1 \dfrac{\ln^n (x) x^{z-1}}{x-1}dx, \Gamma(z)\zeta(z)=\int_0^\infty \dfrac{x^{z-1}}{e^x-1}dx$ the lhs is $\psi_{n} (1/2)$ .at n=1729 we use legendre duplication and differentiate it continously to get the identity $2^{n+1} \psi_n (2z)=\psi_n(z)+\psi_n (z+1/2)\to \psi_n (1/2)=(2^{n+1}-1)\psi_n(1)=(-1)^{n+1} (2^{n+1}-1) \Gamma(n+1)\zeta(n+1)$ we can use the aforementioned integrals to get the result $\int_0^1 \dfrac{\ln^n (x) }{(x-1)\sqrt{x}}dx=(-1)^{n+1}(2^{n+1}-1)\int_0^\infty \dfrac{x^{n}}{e^x-1}dx$ putting n=1729 we get the answer.