How can we prove this integral? $$\int _0^1\int _0^1\int _0^1\frac{\ln ^n\left(xyz\right)}{1-xyz}dxdydz=\frac{\left(-1\right)^n}{2}\Gamma \left(n+3\right)\zeta \left(n+3\right)$$
I have received a solution for above Integral from one of my friends madam Rana Ranino, she also introduce a new Integral transformation(triple Integral to single Integral) like below: $$\int _0^1\int _0^1\int _0^1f\left(xyz\right)dxdydz=\frac{1}{2}\int _0^1f\left(x\right)\ln ^2\left(x\right)dx$$
Prove of transformation: $$\int _0^1\int _0^1\int _0^1f\left(xyz\right)dxdydz=\int _0^1\frac{1}{z}\int _0^1\frac{1}{y}\int _0^{xz}f\left(t\right)dtdxdz$$ $$=\int _0^1\frac{1}{z}\left[\ln x\int _0^{xz}f\left(t\right)dt\right]_{x=0}^{x=1}dz-\int _0^1\ln x\int _0^1f\left(xz\right)dxdz$$ $$=-\int _0^1\ln x\int _0^1f\left(xz\right)dxdz=\int _0^1\frac{\ln x}{x}\int _0^xf\left(t\right)dtdz$$ $$=-\left[\frac{1}{2}\ln ^2x\int _0^xf\left(t\right)dt\right]_{x=0}^{x=1}+\frac{1}{2}\int _0^1f\left(x\right)\ln ^2xdx$$ $$=\frac{1}{2}\int _0^1f\left(x\right)\ln ^2xdx$$
Solution: Now we going to solve above problem with Rana Ranino's transformation: $$\int _0^1\int _0^1\int _0^1\frac{\ln ^n\left(xyz\right)}{1-xyz}dxdydz=\frac{1}{2}\int _0^1\frac{\ln ^{n+2}\left(x\right)}{1-x}dx$$ $$=\frac{1}{2}\sum _{m=0}^{\infty }\int _0^1x^m\ln ^{n+2}xdx=\frac{1}{2}\left(-1\right)^{\left(n+2\right)}\sum _{m=0}^{\infty }\frac{\Gamma \left(n+3\right)}{\left(m+1\right)^{n+3}}$$ $$=\frac{1}{2}\left(-1\right)^n\Gamma \left(n+3\right)\zeta \left(n+3\right)$$
