<s>[INST] Integrating $ \int_0^\infty \frac{x^5}{e^x+1} \, dx $ This improper integral has stumped me and not many integration problems give me problems. However, this one made me think over my limit. Finally, as I could not even get started on this problem, I thought to post it here, and have some talented mathematicians look at it. The problem is to integrate:
$$  \int_0^\infty \frac{x^5}{e^x+1} \, dx $$
 [/INST] $$I=\int _{ 0 }^{ \infty }{ \cfrac { { x }^{ 5 } }{ { e }^{ x }+1 } dx= } \int _{ 0 }^{ \infty }{ \cfrac { { e }^{ -x } }{ { 1+e }^{ -x } } { x }^{ 5 }dx }  =\int _{ 0 }^{ \infty }{ \left( \sum _{ r=1 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx } } { x }^{ 5 } \right) dx } \\ =\sum _{ r=1 }^{ \infty }{ \left\{ \int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } \right\} }$$ 
Let $rx=t$:
$$\int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } ={ \left( -1 \right) }^{ r-1 }\int _{ 0 }^{ \infty }{ { e }^{ -t }{ \left( \cfrac { t }{ r } \right) }^{ 6-1 }\cfrac { dt }{ r } }  =\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \Gamma (6)=120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \\ \\ \therefore I=\sum _{ r=1 }^{ \infty }{ 120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } } =120\eta \left( 6 \right) =120\left\{ \left( 1-{ 2 }^{ 1-6 } \right) \right\} \zeta \left( 6 \right) \\ \\ =120\times \cfrac { 31 }{ 32 } \zeta \left( 6 \right) =\cfrac { 465 }{ 4 } \zeta \left( 6 \right)$$ 
$$\zeta \left( 6 \right) =\zeta \left( 2\times 3 \right) =\cfrac { { \left( -1 \right) }^{ 3+1 }{ B }_{ 6 }{ \left( 2\pi \right) }^{ 6 } }{ 2\left( 6 \right) ! } =\cfrac { { \pi }^{ 6 } }{ 945 }$$ 
Note that $B_6$ refers to the $6$th Bernoulli Number $\left(B_6=\frac{1}{42}\right)$:
$$\therefore I=\cfrac { 465 }{ 4 } \times \cfrac { { \pi }^{ 6 } }{ 945 } =\cfrac { { 31\pi }^{ 6 } }{ 252 }$$
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