What is comma doing here?

$\large{-\displaystyle \int^{\infty}_{0}\frac{x^3}{e^{3x}(1-e^{-x})}dx}$

$\large{\Rightarrow - \displaystyle \int^{\infty}_{0}x^{3} \displaystyle \sum^{\infty}_{r=1}e^{-(r+2)x} dx}$

Put $(r+2)x=t$

$\large{\Rightarrow - \displaystyle \int^{\infty}_{0} t^{3}e^{-t} dt \sum^{\infty}_{r=1} \frac{1}{(r+2)^{4}}}$

Recall gamma function

$\large{\Rightarrow - \Gamma(4) \left( \displaystyle \sum^{\infty}_{r=1} \frac{1}{r^{4}}-1-\frac{1}{2^4} \right)}$

$\Large{=3! \left(\frac{17}{16}- \zeta(4) \right)}$

$\Large{\boxed{\frac{51}{8}-\frac{\pi^4}{15}}}$

$I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ 3 } }{ { e }^{ 2x }-{ e }^{ 3x } } dx } \\ \quad =\int _{ 0 }^{ \infty }{ \frac { { { e }^{ -3x }x }^{ 3 } }{ { e }^{ -x }-1 } dx }$

Now we substitute: ${ e }^{ -x }=x$

$\therefore \quad I=\int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 }{ \left( lnx \right) }^{ 3 } }{ x-1 } } \\ \quad \quad \quad \quad ={ \psi }_{ 3 }\left( 3 \right)$

We use the relation: ${ \psi }_{ n }\left( z \right) ={ \left( -1 \right) }^{ n+1 }n!\zeta \left( n+1,z \right)$

Here $\zeta \left( n+1,z \right)$ is Hurwitz Zeta function.

$\therefore \quad I=-3!\zeta \left( 4,3 \right) \\ \quad \quad \quad =-3!\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( n+3 \right) }^{ 4 } } } \\ \quad \quad \quad =\frac { {- \pi }^{ 4 } }{ 15 } +\frac { 51 }{ 8 }$