Because $$-\int_0^\infty \frac{e^{-zt}}{t^z}dt=z^2\Gamma(-z),$$ holds for $0<\Re z<1$ then using the change of variable $t=nx$ one has that $$-\frac{1}{n^{z-1}}\int_0^\infty \frac{e^{-znx}dx}{x^z}=z^2\Gamma(-z),$$ and multiplying by $\frac{1}{n^3}$ for integers $n\geq 1$ and taking the sum, one has if there were not mistakes $$\zeta(3)=-\frac{-1}{z^2\Gamma(-z)}\int_0^\infty\frac{1}{x^z}\left(\sum_{n=1}^\infty\frac{e^{-znx}}{n^{z+2}}\right)dx.$$ Then I tried know if it was right. First one knows, using Wolfram Alpha, that the second factor in previous integrand is $Li_{z+2}(e^{-xz})$.
Question. Were rights my calculations? Can you integrating $$\int_0^\infty x^{-z}Li_{z+2}(e^{-xz})dx$$ show that it is equal to $$-\zeta(3)z^2\Gamma(-z)?$$
Thanks in advance.
I am not able to prove (previous directly as I am asking) with Wolfram Alpha online calculator, with this my code and with standard computation time:
int x^(-z) PolyLog(z+2,e^(-xz)) dx from x=0 to infinite.