[Largeint_{0}^{1}frac{ln^{2}left ( x
ight )mathrm{Li}_{2}left ( x
ight )}{1-x}mathrm{d}x=-11zeta left ( 5
ight )+6zeta left ( 3
ight )zeta left ( 2
ight )
]
(Largemathbf{Proof:})
[egin{align*}
int_{0}^{1}frac{ln^{2}left ( x
ight )mathrm{Li}_{2}left ( x
ight )}{1-x}mathrm{d}x &=int_{0}^{1}frac{ln^{2}left ( x
ight )}{1-x}sum_{n=1}^{infty }frac{x^{n}}{n^{2}}mathrm{d}x \
&=int_{0}^{1}frac{ln^{2}left ( x
ight )}{1-x}left [ sum_{n=1}^{infty }frac{1}{n^{2}}-sum_{n=1}^{infty }frac{1-x^{n}}{n^{2}}
ight ]mathrm{d}x \
&=zeta left ( 2
ight )int_{0}^{1}frac{ln^{2}left ( x
ight )}{1-x}mathrm{d}x-int_{0}^{1}frac{ln^{2}left ( x
ight )}{1-x}sum_{n=1}^{infty }frac{1-x^{n}}{n^{2}}mathrm{d}x
end{align*}]
However,
[egin{align*}
int_{0}^{1}frac{ln^{2}left ( x
ight )}{1-x}mathrm{d}x&=int_{0}^{1}lnleft ( 1-x
ight )left [ 2lnleft ( x
ight )frac{1}{x}
ight ]mathrm{d}x=-2int_{0}^{1}mathrm{Li}{}'_{2}left ( x
ight )lnleft ( x
ight )mathrm{d}x \
&=2int_{0}^{1}mathrm{Li}_{2}left ( x
ight )frac{1}{x}mathrm{d}x=2int_{0}^{1}mathrm{Li}{}'_{3}left ( x
ight )mathrm{d}x=2mathrm{Li}_{3}left ( 1
ight )=2zeta left ( 3
ight )
end{align*}]
Such that
[int_{0}^{1}frac{ln^{2}left ( x
ight )mathrm{Li}_{2}left ( x
ight )}{1-x}mathrm{d}x=2zeta left ( 2
ight )zeta left ( 3
ight )-int_{0}^{1}frac{ln^{2}left ( x
ight )}{1-x}sum_{n=1}^{infty }frac{1-x^{n}}{n^{2}}mathrm{d}x
]
Also
[egin{align*}
int_{0}^{1}frac{ln^{2}left ( x
ight )}{1-x}sum_{n=1}^{infty }frac{1-x^{n}}{n^{2}}mathrm{d}x &=sum_{n=1}^{infty }frac{1}{n^{2}}int_{0}^{1}ln^{2}left ( x
ight )frac{1-x^{n}}{1-x}mathrm{d}x \
&=sum_{n=1}^{infty }frac{1}{n^{2}}int_{0}^{1}ln^{2}left ( x
ight )sum_{k=1}^{n}x^{k-1}mathrm{d}x \
&= sum_{n=1}^{infty }frac{1}{n^{2}}sum_{k=1}^{n}underset{=displaystyle frac{2}{k^{3}}}{underbrace{int_{0}^{1}ln^{2}left ( x
ight )x^{k-1}mathrm{d}x}}=2sum_{n=1}^{infty }frac{H_{n}^{left ( 3
ight )}}{n^{2}}
end{align*}]
The last sum can be evaluated with the generating function(displaystyle sum_{n=1}^{infty }x^{n}H_{n}^{left ( 3 ight )}=frac{mathrm{Li}_{3}left ( x ight )}{1-x}).Namely
[egin{align*}
sum_{n=1}^{infty }frac{x^{n}}{n}H_{n}^{left ( 3
ight )}&=int_{0}^{x}frac{mathrm{Li}_{3}left ( t
ight )}{t}mathrm{d}t+int_{0}^{x}frac{mathrm{Li}_{3}left ( t
ight )}{1-t}mathrm{d}t \
&=mathrm{Li}_{4}left ( x
ight )-lnleft ( 1-x
ight )mathrm{Li}_{3}left ( x
ight )+int_{0}^{x}lnleft ( 1-t
ight )mathrm{Li}{}'_{3}left ( t
ight )mathrm{d}t\
&=mathrm{Li}_{4}left ( x
ight )-lnleft ( 1-x
ight )mathrm{Li}_{3}left ( x
ight )+int_{0}^{x}lnleft ( 1-t
ight )frac{mathrm{Li}_{2}left ( t
ight )}{t}mathrm{d}t \
&=mathrm{Li}_{4}left ( x
ight )-lnleft ( 1-x
ight )mathrm{Li}_{3}left ( x
ight )+int_{0}^{x}mathrm{Li}_{2}left ( t
ight )mathrm{Li}{}'_{2}left ( t
ight )mathrm{d}t \
&=mathrm{Li}_{4}left ( x
ight )-lnleft ( 1-x
ight )mathrm{Li}_{3}left ( x
ight )-frac{1}{2}mathrm{Li}_{2}^{2}left ( x
ight )
end{align*}]
[egin{align*}
sum_{n=1}^{infty }frac{H_{n}^{left ( 3
ight )}}{n^{2}}&=int_{0}^{1}frac{mathrm{Li}_{4}left ( t
ight )}{t}mathrm{d}t-int_{0}^{1}frac{lnleft ( 1-t
ight )mathrm{Li}_{3}left ( t
ight )}{t}mathrm{d}t-frac{1}{2}int_{0}^{1}frac{mathrm{Li}_{2}^{2}left ( t
ight )}{t}mathrm{d}t \
&= zeta left ( 5
ight )+mathrm{Li}_{2}left ( 1
ight )mathrm{Li}_{3}left ( 1
ight )-int_{0}^{1}mathrm{Li}_{2}left ( t
ight )frac{mathrm{Li}_{2}left ( t
ight )}{t}mathrm{d}t-frac{1}{2}int_{0}^{1}frac{mathrm{Li}_{2}^{2}left ( t
ight )}{t}mathrm{d}t\
&=zeta left ( 5
ight )+zeta left ( 2
ight )zeta left ( 3
ight ) -frac{3}{2}int_{0}^{1}frac{mathrm{Li}_{2}^{2}left ( t
ight )}{t}mathrm{d}t
end{align*}]
The last integral integrate once by parts,we find
[egin{align*}
I &=int_{0}^{1}frac{mathrm{Li}_{2}^{2}left ( x
ight )}{x}mathrm{d}x=int_{0}^{1}mathrm{Li}_{2}left ( x
ight )mathrm{d}left ( mathrm{Li}_{3}left ( x
ight )
ight ) \
&=mathrm{Li}_{2}left ( 1
ight )mathrm{Li}_{3}left ( 1
ight )+int_{0}^{1}frac{mathrm{Li}_{3}left ( x
ight )lnleft ( 1-x
ight )mathrm{d}x}{x} \
&=zeta left ( 2
ight ) zeta left ( 3
ight )-sum_{n=1}^{infty }sum_{k=1}^{infty }frac{1}{n^{3}kleft ( n+k
ight )}
end{align*}]
where the last line is obtained by expanding (displaystyle mathrm{Li}_{3}left ( x
ight )) and (lnleft ( 1-x
ight )) into series and integrating. Now if we donote(displaystyle H_{n}=sum_{k=1}^{n}frac{1}{k}) the (n)th harmonic number, the sum with respect to (k) can be written as$$sum_{k=1}^{infty }frac{1}{kleft ( n+k
ight )}=frac{1}{n}sum_{k=1}^{infty }left ( frac{1}{k}-frac{1}{n+k}
ight )=frac{H_{n}}{n}$$
so that$$I=zeta left ( 2
ight )zeta left ( 3
ight )-sum_{n=1}^{infty }frac{H_{n}}{n^{4}}=2zeta left ( 2
ight )zeta left ( 3
ight )-3zeta left ( 5
ight )$$
so we have
[sum_{n=1}^{infty }frac{H_{n}^{left ( 3
ight )}}{n^{2}}=frac{11}{2}zeta left ( 5
ight )-2left ( 2
ight )zeta left ( 3
ight )
]
[Rightarrow int_{0}^{1}frac{ln^{2}left ( x
ight )}{1-x}sum_{n=1}^{infty }frac{1-x^{n}}{n^{2}}mathrm{d}x=11zeta left ( 5
ight )-4zeta left ( 2
ight )zeta left ( 3
ight )
]
Now we have the final result:
[oxed{displaystyle color{blue}{int_{0}^{1}frac{ln^{2}left ( x
ight )mathrm{Li}_{2}left ( x
ight )}{1-x}mathrm{d}x=-11zeta left ( 5
ight )+6zeta left ( 3
ight )zeta left ( 2
ight )}}
]