What's about $\int_0^\infty x^{-z}Li_{z+2}(e^{-xz})dx$ as $-\zeta(3)z^2\Gamma(-z)$? 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.


First, we use the definition of the polylogarithm function 
$$\text{Li}_{s}(z)\equiv \sum_{n=1}^\infty\frac{z^n}{n^s}$$
to write 
$$\text{Li}_{z+2}(e^{-xz})=\sum_{n=1}^\infty\frac{e^{-nxz}}{n^{z+2}}$$
Then, proceeding formally, without regard for rigor, we have  
$$\begin{align}
\int_0^\infty x^{-z}\text{Li}_{z+2}(e^{-xz})\,dx&=\sum_{n=1}^\infty \frac{1}{n^{z+2}}\int_0^\infty x^{-z}e^{-nxz}\,dx\\\\
&=\sum_{n=1}^\infty \frac{1}{n^{z+2}}\int_0^\infty \left(\frac{t}{nz}\right)^{-z}e^{-t}\,\frac{1}{nz}\,dt\\\\
&=z^{z-1}\sum_{n=1}^\infty \frac{1}{n^3}\int_0^\infty t^{-z}e^{-t}\,dt\\\\
&=z^{z-1}\zeta(3)\Gamma(1-z)\\\\
&=-z^z\zeta(3)\Gamma(-z)
\end{align}$$
