He encontrado ahora, pero no es muy elegante:
$$
\int_0^1 \ln\left(K(x)\right)\espacio dx =\int_0^1 \ln\left(x^{-x}\cdot(2\pi)^{-\frac{x}{2}}\cdot e^{\frac{1}{2}x(x+1)-\frac{\gamma}{2}x^2}\cdot\prod_{k=1}^\infty \left[\frac{e^{x+\frac{x^2}{2k}}}{\left(1+\frac{x}{k}\right)^{x+k}}\right]\right)\espacio dx ={-\int_0^1 {x\ln\left(x\right)\espacio dx}}-\int_0^1 \frac x 2\ln\left(2\pi\right)\espacio dx+\int_0^1 \frac 1 2 x(x+1) espacio\dx-\int_0^1 \frac{\gamma}{2}x^2\espacio dx+\int_0^1 \sum_{k=1}^{\infty}\left[x+\frac{x^2}{2k}-(x+k)\ln\left(1+\frac{x}{k}\right)\right]\espacio dx=-\left[\frac {x^2}{2}\ln(x)-\frac{x^2}{4}\right]_{0}^{1}-\left[\frac{x^2}{4}\ln(2\pi)\right]_{0}^{1}+\left[\frac{x^3}{6}+\frac{x^2}{4}\right]_{0}^{1}-\left[\frac{\gamma}{6}x^3\right]_{0}^{1}+\sum_{k=1}^{\infty}\left(\left[\frac{x^2}{2}\right]_0^1+\left[\frac{x^3}{6k}\right]_0^1-\left[\frac{1}{2}(x+k)^2\ln\left(1+\frac{x}{k}\right)-\frac{x}{2}\left(\frac{x}{2}+k\right)\right]_0^1\right)=\frac{1}{4}-\frac{1}{4}\ln(2\pi)+\frac{1}{6}+\frac{1}{4}-\frac{\gamma}{6}+\sum_{k=1}^{\infty}\left(\frac{1}{2}+\frac{1}{6k}-\frac{1}{2}(k+1)^2\ln\left(1+\frac{1}{k}\right)+\frac{1}{2}\left(\frac{1}{2}+k\right)\right)=\frac{2}{3}-\frac{1}{4}\ln(2\pi)-\frac{\gamma}{6}+\sum_{k=1}^{\infty}\left(\frac{3}{4}+\frac{1}{6k}+\frac{k}{2}+\frac{1}{2}\ln\left(\frac{k^{(k+1)^2}}{(k+1)^{(k+1)^2}}\right)\right)
$$
Ahora, si definimos $H_2(n):=\prod_{k=1}^{n} k^{k^2}$ $H_1(n):=K(n+1)=\prod_{k=1}^{n} k^k$ podemos calcular la n-ésima suma parcial de la infinita suma anterior:
$$
S_n=\sum_{k=1}^{n}\left(\frac{3}{4}+\frac{1}{6k}+\frac{k}{2}+\frac{1}{2}\ln\left(\frac{k^{(k+1)^2}}{(k+1)^{(k+1)^2}}\right)\right)=\frac{3}{4}n+\frac{1}{6}H_n+\frac{1}{4}n(n+1)+\frac{1}{2}\ln\left(\prod_{k=1}^{n}\frac{k^{k^2+2k+1}}{(k+1)^{(k+1)^2}}\right)=\frac{3}{4}n+\frac{1}{6}H_n+\frac{1}{4}n(n+1)+\frac{1}{2}\ln\left(\frac{H_2(n)\cdot H_1(n)^2\cdot n!}{\frac{H_2(n+1)}{H_2(1)}}\right)={\frac{3}{4}n+\frac{1}{6}H_n+\frac{1}{4}n(n+1)+\frac{1}{2}\ln\left(\frac{H_1(n)^2\cdot n!}{(n+1)^{(n+1)^2}}\right)}
$$
Ahora, $\lim_{n\to\infty} S_n$ se puede calcular mediante:
$$
A=\lim_{n\to\infty}\frac{1^1\cdot 2^2\cdots n^n}{n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}}\ffi \lim_{n\to\infty}\frac{H_1(n)}{A\cdot n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}}=1\iff\lim_{n\to\infty}\ln\left(\frac{H_1(n)}{A\cdot n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}}\right)=0
$$
$$
\gamma=\lim_{n\to\infty}\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}-\ln(n)\ffi \lim_{n\to\infty}\left(H_n-\ln(n)-\gamma\right)=0
$$
$$
1=\lim_{n\to\infty}\frac{n!}{n^{n+\frac{1}{2}}\cdot e^{-n}\cdot\sqrt{2\pi}}\ffi\lim_{n\to\infty}\ln\left(\frac{n!}{n^{n+\frac{1}{2}}\cdot e^{-n}\cdot\sqrt{2\pi}}\right)=0
$$
Y por lo tanto:
$$
\lim_{n\to\infty} S_n=\lim_{n\to\infty} \left[{\frac{3}{4}n+\frac{1}{6}H_n+\frac{1}{4}n(n+1)+\frac{1}{2}\ln\left(\frac{H_1(n)^2\cdot n!}{(n+1)^{(n+1)^2}}\right)}\right]=\lim_{n\to\infty} \left[{\frac{3}{4}n+\frac{1}{6}\left(\ln(n)+\gamma\right)+\frac{1}{4}n(n+1)+\frac{1}{2}\ln\left(\frac{\left(A\cdot n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}\right)^2\cdot \left(n^{n+\frac{1}{2}}\cdot e^{-n}\cdot\sqrt{2\pi}\right)}{(n+1)^{(n+1)^2}}\right)}\right]=\lim_{n\to\infty} \left[{\frac{\gamma}{6}+\frac{1}{2}\ln\left(\frac{e^{\frac{3}{2}n}\cdot n^{\frac{1}{3}}\cdot e^{\frac{1}{2}n^2+\frac{1}{2}n}\cdot A^2\cdot n^{n^2+n+\frac{1}{6}}\cdot e^{-\frac{n^2}{2}}\cdot n^{n+\frac{1}{2}}\cdot e^{-n}\cdot\sqrt{2\pi}}{(n+1)^{n^2+2n+1}}\right)}\right]=\lim_{n\to\infty} \left[{\frac{\gamma}{6}+\ln(A)+\frac{1}{4}\ln(2\pi)+\frac{1}{2}\ln\left(\frac{e^{n}\cdot n^{n^2+2n+1}}{(n+1)^{n^2+2n+1}}\right)}\right]=\frac{\gamma}{6}+\ln(A)+\frac{1}{4}\ln(2\pi)+\frac{1}{2}\lim_{n\to\infty} \left[n-(n^2+2n+1)\cdot \ln\left(1+\frac{1}{n}\right)\right]=\frac{\gamma}{6}+\ln(A)+\frac{1}{4}\ln(2\pi)+\frac{1}{2}\lim_{n\to\infty} \left[n-(n^2+2n+1)\cdot \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k\cdot n^k}\right]=\frac{\gamma}{6}+\ln(A)+\frac{1}{4}\ln(2\pi)+\frac{1}{2}\lim_{n\to\infty} \left[n-\left(n-\frac{1}{2}+\sum_{k=3}^{\infty}\frac{(-1)^{k+1}}{k\cdot n^{k-2}}\right)-\left(2+2\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k\cdot n^{k-1}}\right)- \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k\cdot n^k}\right]=\frac{\gamma}{6}+\ln(A)+\frac{1}{4}\ln(2\pi)+\frac{1}{2}\lim_{n\to\infty} \left[-\frac{3}{2}-\sum_{k=3}^{\infty}\frac{(-1)^{k+1}}{k\cdot n^{k-2}}-2\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k\cdot n^{k-1}}- \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k\cdot n^k}\right]=\frac{\gamma}{6}+\ln(A)+\frac{1}{4}\ln{(2\pi)}+\frac{1}{2}\left[-\frac{3}{2}\right]=\frac{\gamma}{6}+\ln(A)+\frac{1}{4}\ln(2\pi)-\frac{3}{4}
$$
Y por lo tanto:
$$
\int_0^1 \ln\left(K(x)\right)\espacio dx=\frac{2}{3}-\frac{1}{4}\ln(2\pi)-\frac{\gamma}{6}+\sum_{k=1}^{\infty}\left(\frac{3}{4}+\frac{1}{6k}+\frac{k}{2}+\frac{1}{2}\ln\left(\frac{k^{(k+1)^2}}{(k+1)^{(k+1)^2}}\right)\right)=\frac{2}{3}-\frac{1}{4}\ln(2\pi)-\frac{\gamma}{6}+\frac{\gamma}{6}+\ln(A)+\frac{1}{4}\ln(2\pi)-\frac{3}{4}=\ln(A)-\frac{1}{12}=-\zeta'(-1)
$$
Pero todavía me pregunto si hay una manera más rápida para evaluar la integral, así que por favor, si usted encuentra una elegante prueba, sería muy lindo post!
