How to evaluate the following interesting integral?

Show that the integral,

$$\frac{3}{x^3}\int_{0}^{x}\frac{t^3}{e^t-1}=\frac{\pi^4}{5x^3}+ O(e^{-x}) $$ for $x>>1$

I tried ignoring the one in the denominator and then trying by parts but it gets too long and is maybe wrong.

Also, tried approximating the upper limit as infinity but that doesn't help and is probably wrong.

Note as Hans spotted we have $$\frac{t^3e^{-t}}{1-e^{-t}} = t^3e^{-t}(\sum_{j=0}^{\infty}e^{-nt}) = \sum_{j=1}^{\infty}t^3e^{-jt}$$

whenever $t > 0$.

By integration by parts (no I totally didn't use wolfram alpha)

$$\int_{a}^{x}t^3e^{-jt}dt = \frac{e^{-aj} (a^3 j^3 + 3 a^2 j^2 + 6 a j + 6)}{j^4} - \frac{e^{-j x} (j^3 x^3 + 3 j^2 x^2 + 6 j x + 6)}{j^4}$$

Now whenever $0<a<x$ we have

$$\int_{a}^{x}\frac{t^3e^{-t}}{1-e^{-t}} dt= \sum_{j=1}^{\infty}\left(\frac{e^{-aj} (a^3 j^3 + 3 a^2 j^2 + 6 a j + 6)}{j^4} - \frac{e^{-j x} (j^3 x^3 + 3 j^2 x^2 + 6 j x + 6)}{j^4}\right)$$

Now set $b > 0$

we have

$$\sum_{j=1}^{\infty}\frac{e^{-bj} (b^3 j^3 + 3 b^2 j^2 + 6 b j + 6)}{j^4} = b^{3}\sum_{j=1}^{\infty}\frac{e^{-bj}}{j}+3b^2\sum_{j=1}^{\infty}\frac{e^{-bj}}{j^2}+6b\sum_{j=1}^{\infty}\frac{e^{-bj}}{j^3}+6\sum_{j=1}^{\infty}\frac{e^{-bj}}{j^4}$$

Now note

$$b^{3}\sum_{j=1}^{\infty}\frac{e^{-bj}}{j} = -b^3 \log(1-e^{-b}) \rightarrow 0$$

as $b \rightarrow 0^{+}$.

$$3b^2\sum_{j=1}^{\infty}\frac{e^{-bj}}{j^2} \leq 3b^2 \zeta(2) \rightarrow 0$$

as $b \rightarrow 0^{+}$.

$$6b\sum_{j=1}^{\infty}\frac{e^{-bj}}{j^3} \leq 6b\zeta(3) \rightarrow 0$$

as $b \rightarrow 0^{+}$.

$$6\sum_{j=1}^{\infty}\frac{e^{-bj}}{j^4} \rightarrow 6\zeta(4)$$

as $b \rightarrow 0^{+}$.

Thus

$$\int_{0}^{x}\frac{t^3e^{-t}}{1-e^{-t}} dt = \lim_{a \rightarrow 0^{+}}\int_{a}^{x}\frac{t^3e^{-t}}{1-e^{-t}} dx = 6\zeta(4)-\sum_{j=1}^{\infty}\frac{e^{-j x} (j^3 x^3 + 3 j^2 x^2 + 6 j x + 6)}{j^4}$$

Thus for $x > 1$ we have

$$|\int_{0}^{x}\frac{t^3e^{-t}}{1-e^{-t}} dt-6\zeta(4)| \leq 18x^3\sum_{j=1}^{\infty}e^{-jx} = 18x^3\frac{e^{-x}}{1-e^{-x}} \leq \frac{18}{1-e^{-1}}e^{-x}x^{3}$$

This proves your result when we multiply both sides by $\frac{3}{x^3}$ and use $\zeta(4) = \frac{\pi^4}{90}$.