Seven seas in this world!

I used the relation:

$\zeta(s) \Gamma(s)=\large\int_{0}^{\infty}{ \dfrac{x^{s-1}}{e^{x}-1} \, dx}$

Therefore

$-\zeta(8)\Gamma(8)=\large\int_0^\infty \dfrac{x^{7}}{1-e^{x}} \, dx$

LHS:

We can first start by evaluating $\zeta(8)$ :

We use the relation:

$\pi^{\dfrac{-s}{2}} \Gamma\left(\dfrac{s}{2}\right) \zeta(s)= \pi^{-\dfrac{1-s}{2}} \Gamma\left(\dfrac{1-s}{2}\right) \zeta\left(1-s\right)$

$\implies \zeta(s)=\dfrac{\pi^{s-\dfrac{1}{2}} \Gamma\left(\dfrac{1-s}{2}\right)\zeta\left(1-s\right)}{\Gamma\left(\dfrac{s}{2}\right)}$

$\zeta(8)=\dfrac{\pi^{\dfrac{15}{2}} \Gamma\left(\dfrac{-7}{2}\right)\zeta(-7)}{\Gamma(4)}$

$\zeta(8)=\dfrac{\pi^{\dfrac{15}{2}} \Gamma\left(\dfrac{-7}{2}\right)\dfrac{\beta_8}{8}}{\Gamma(4)}$

$\zeta(8)=\dfrac{\pi^{\dfrac{15}{2}} \Gamma\left(\dfrac{-7}{2}\right)\dfrac{1}{240}}{\Gamma(4)}$

$\zeta(8)=\dfrac{\pi^{\dfrac{15}{2}} \Gamma\left(\dfrac{-7}{2}\right)}{3! 240}$

$\zeta(8)=\dfrac{\pi^{\dfrac{15}{2}} \Gamma\left(\dfrac{-7}{2}\right)}{1440}$

We now use Euler’s reflection formula to compute $\Gamma\left(\dfrac{-7}{2}\right)$

$\Gamma(1-z) \Gamma(z)=\dfrac{\pi}{\sin \pi z}$

$\Gamma\left(\dfrac{9}{2}\right) \Gamma\left(\dfrac{-7}{2}\right)=\dfrac{\pi}{\sin -\dfrac{7 \pi}{2}}$

$\Gamma\left(\dfrac{9}{2}\right) \Gamma\left(\dfrac{-7}{2}\right)=\dfrac{\pi}{\sin \dfrac{\pi}{2}}$

$\Gamma\left(\dfrac{9}{2}\right) \Gamma\left(\dfrac{-7}{2}\right)=\pi$

We can then evaluate $\Gamma\left(\dfrac{9}{2}\right)$ separately, we can reduce it using the fact that:

$\Gamma(s+1)=s\Gamma(s)$

$\therefore \Gamma\left(\dfrac{9}{2}\right)=\dfrac{105}{16} \Gamma\left(\dfrac{1}{2}\right)$

We can then compute $\Gamma\left(\dfrac{1}{2}\right)$ using:

$\Gamma(s)=\int_{0}^{\infty} t^{s-1} e^{-t} \, dt$

$\Gamma\left(\dfrac{1}{2}\right)=\int_{0}^{\infty} t^{\dfrac{1}{2}} e^{-t} \, dt$

Let $t=x^{2}$ , the variable $x$ is independent to the one in the question. Thus $dt=2xdx$

$= 2 \int_{0}^{\infty} x^{-1} e^{-x^{2}} x \, dx$

$= 2 \int_{0}^{\infty} e^{-x^{2}} \, dx$

Since the function we are integrating is an even function, thus instead of having the $2$ in front of the integral we can re-write it as:

$= \int_{-\infty}^{\infty} {e^{-x^{2}} \, dx}$

We can let the above integral be $I$ then square it, change the variable of integration on the other, since it won’t affect its value then evaluate.

$=\left(\int_{-\infty}^{\infty} {e^{-x^{2}} \, dx}\right)\left(\int_{-\infty}^{\infty} {e^{-y^{2}} \, dy}\right)$

$= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} {e^{-(x^{2}+y^{2})} \, dxdy}$

Let $x=r\cos \theta$ and $y=r\sin \theta$ . Computing the Jacobian yields $|J|=r$

$= \int_0^{2\pi} \int_0^\infty r e^{-r^{2}} \, dr d\theta$

Let $u=r^{2}$ , $\dfrac{1}{2} du= rdr$ .

$=\dfrac{1}{2} \int_{0}^{2\pi} \int_{0}^{\infty} {e^{-u} \, du d\theta}$

$=\dfrac{1}{2} \int_0^{2\pi} -e^{-u}|_{0}^{\infty} \, d\theta$

$=\dfrac{1}{2} \int_0^{2\pi} -e^{-u}|_{0}^{\infty} \, d\theta$

$=\dfrac{1}{2} \int_{0}^{2\pi} \, d\theta$

$=\pi$

$\therefore I^{2}=\pi$ meaning $I=\sqrt{\pi}$

Therefore $\Gamma \left(\dfrac{1}{2} \right) = \sqrt{\pi}$

This means that $\Gamma \left(\dfrac{9}{2}\right) = \dfrac{105\sqrt{\pi}}{16}$

Since:

$\Gamma\left(\dfrac{9}{2}\right) \Gamma\left(\dfrac{-7}{2}\right)=\pi$

This means that:

$\Gamma\left(\dfrac{-7}{2}\right)=\dfrac{16\sqrt{\pi}}{105}$

Then:

$\zeta(8)=\dfrac{\pi^{\dfrac{15}{2}} \dfrac{16\sqrt{\pi}}{105}}{1440}$

$\therefore \zeta(8) = \dfrac{\pi^{8}}{9450}$

Finally we reach our answer to the integral to be:

$\large\int_0^\infty \dfrac{x^{7}}{1-e^{x}} \, dx = -\zeta(8)\Gamma(8) = -\dfrac{\pi^{8}}{9450} \left(8-1 \right)!= -\dfrac{8\pi^{8}}{15}$

$\therefore A+B+C= 8+15+8= 31$

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

We get: $I=\int _{ 0 }^{ 1 }{ \frac { { \left( lnx \right) }^{ 7 } }{ x-1 } }$

We will use poly-gamma function for this.

${ \psi }_{ 7 }\left( n+1 \right) =\int _{ 0 }^{ 1 }{ \frac { { x }^{ n }{ \left( lnx \right) }^{ 7 } }{ x-1 } }$

Now, we substitute $n=0$

${ \psi }_{ 7 }\left( 1 \right) =\int _{ 0 }^{ 1 }{ \frac { { \left( lnx \right) }^{ 7 } }{ x-1 } }$

Now, 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.

Now, we substitute $n=7$ and $z=1$ .

${ \psi }_{ 7 }\left( 1 \right) =-7!\zeta \left( 8,1 \right)$

Now, we use the property: $\zeta \left( n+1,1 \right) =\zeta \left( n+1 \right)$

Therefore we get: ${ \psi }_{ 7 }\left( 1 \right) =-7!\zeta \left( 8 \right)$

Therefore the answer is: $\frac{-8{\pi}^{8}}{15}$

Note that $\zeta \left( 8 \right)=\frac{{\pi}^{8}}{9450}$