So I was studying the following function:
$f(x)=\frac{x^{s-1}}{s^{x}-1}$ where $s$ is any natural number greater than $1$ by playing with Wolfram-Alpha a bit I observed that by taking the integral of $f(x)$ from $0$ to $\infty$ or $\int_{0}^{{\infty}}\frac{x^{s-1}}{{s}^{x}-1}$ and setting $s=2n+1$
We get the following results provided by Wolfram:
$$\int_{0}^{{\infty}}\frac{x^{3-1}}{{3}^{x}-1}=2\frac{\zeta(3)}{\ln^{3}(3)}$$
$$\int_{0}^{{\infty}}\frac{x^{5-1}}{{5}^{x}-1}=24\frac{\zeta(5)}{\ln^{5}(5)}$$
$$\int_{0}^{{\infty}}\frac{x^{7-1}}{{7}^{x}-1}=720\frac{\zeta(7)}{\ln^{7}(7)}$$
. . . And so on up to some odd numbers, I know that there is no elementary way of evaluating this integral but I wanted to see a hindsight of how this results came about and how is it related to the Riemann-Zeta function!
With variable change ${n^x=e^t}$
\begin{align} I(n)=\int_{0}^{{\infty}}\frac{x^{n-1}}{{n}^{x}-1}dx= &\>\frac1{\ln^{n}n} \int_{0}^{{\infty}}\frac{t^{n-1}}{e^{t}-1}dt\\ = &\>\frac1{\ln^{n}n} \sum_{k=1}^\infty\int_{0}^{{\infty}}t^{n-1} e^{-kt}dt\\ = &\>\frac{(n-1)!}{{\ln^{n}n}} \sum_{k=1}^\infty\frac1{k^{n}}\\ = &\>\frac{(n-1)!}{\ln^{n}n} \zeta(n) \end{align}
which produces
$$I(3)=\int_{0}^{{\infty}}\frac{x^{3-1}}{{3}^{x}-1}dx=2\frac{\zeta(3)}{\ln^{3}3}$$
$$I(5)=\int_{0}^{{\infty}}\frac{x^{5-1}}{{5}^{x}-1}dx=24\frac{\zeta(5)}{\ln^{5}5}$$
$$I(7)=\int_{0}^{{\infty}}\frac{x^{7-1}}{{7}^{x}-1}dx=720\frac{\zeta(7)}{\ln^{7}7}$$