复杂的对数积分（二）

[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 ) ]

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(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 )}} ]