[Largedisplaystyle int_{0}^{frac{pi }{2}}x^{2}lnleft ( sin x
ight )lnleft ( cos x
ight )mathrm{d}x
]
(Largemathbf{Solution:})
Tools Needed
[frac{1}{kleft ( j- k
ight )^{2}}=frac{1}{j^{2}k}-frac{1}{j^{2}left ( k- j
ight )}+frac{1}{jleft ( k- j
ight )^{2}}
]
[frac{1}{kleft ( j+ k
ight )^{2}}=frac{1}{j^{2}k}-frac{1}{j^{2}left ( k+ j
ight )}-frac{1}{jleft ( k+j
ight )^{2}}
]
[lnleft ( sin x
ight )=-ln 2-sum_{k=1}^{infty }frac{cosleft ( 2kx
ight )}{k}
]
[lnleft ( cos x
ight )=-ln 2-sum_{k=1}^{infty }left ( -1
ight )^{k}frac{cosleft ( 2kx
ight )}{k}
]
[cosleft ( 2jx
ight )cosleft ( 2kx
ight )=frac{1}{2}left [ cosleft ( 2left ( j-k
ight )x
ight )+cosleft ( 2left ( j+k
ight )x
ight )
ight ]
]
[int_{0}^{frac{pi }{2}}x^{2}cosleft ( 2kx
ight )mathrm{d}x=egin{cases}
left ( -1
ight )^{k}displaystyle frac{pi }{4k^{2}}& ext{ if } k
eq 0 \
displaystyle frac{pi ^{3}}{24}& ext{ if } k=0
end{cases}]
Tool Use
[egin{align*}
&int_{0}^{frac{pi }{2}}x^{2}lnleft ( sin x
ight )lnleft ( cos x
ight )mathrm{d}x \
&=int_{0}^{frac{pi }{2}}x^{2}left ( ln 2+sum_{k=1}^{infty }frac{cosleft ( 2kx
ight )}{k}
ight )left ( ln 2+sum_{k=1}^{infty }left ( -1
ight )^{k}frac{cosleft ( 2kx
ight )}{k}
ight )mathrm{d}x \
&=ln^{2}2 int_{0}^{frac{pi }{2}}x^{2}mathrm{d}x+ln 2sum_{k=1}^{infty }frac{1}{k}int_{0}^{frac{pi }{2}}x^{2}cosleft ( 4kx
ight )mathrm{d}x\
&~~~+sum_{j=1}^{infty }sum_{k=1}^{infty }frac{left ( -1
ight )^{k}}{2jk}int_{0}^{frac{pi }{2}}x^{2}left [ cosleft ( 2left ( j-k
ight )x
ight )+cosleft ( 2left ( j+k
ight )x
ight )
ight ]mathrm{d}x \
&=frac{pi ^{3}}{24}ln^{2}2+ln 2frac{pi }{16}zeta left ( 3
ight ) \
&~~~+frac{pi }{8}sum_{j=1}^{infty }frac{left ( -1
ight )^{j}}{j}sum_{k=1}^{j-1}frac{1}{kleft ( j-k
ight )^{2}}+frac{pi }{8}sum_{j=1}^{infty }frac{left ( -1
ight )^{j}}{j^{2}}frac{pi ^{2}}{6}+frac{pi }{8}sum_{j=1}^{infty }frac{left ( -1
ight )^{j}}{j}sum_{k=j+1}^{infty }frac{1}{kleft ( j-k
ight )^{2}} \
&~~~+frac{pi }{8}sum_{j=1}^{infty }frac{left ( -1
ight )^{j}}{j}sum_{k=1}^{infty }frac{1}{kleft ( j+k
ight )^{2}} \
&=frac{pi ^{3}}{24}ln^{2}2+ln 2frac{pi }{16}zeta left ( 3
ight ) \
&~~~+frac{pi }{8}sum_{j=1}^{infty }frac{left ( -1
ight )^{j}}{j}left ( frac{2}{j^{2}}H_{j-1}+frac{1}{j}H_{j-1}^{left ( 2
ight )}
ight )-frac{pi ^{5}}{576}+frac{pi }{8}sum_{j=1}^{infty }frac{left ( -1
ight )^{j}}{j}left ( -frac{1}{j^{2}}H_{j}+frac{1}{j}frac{pi ^{2}}{6}
ight )\
&~~~+frac{pi }{8}sum_{j=1}^{infty }frac{left ( -1
ight )^{j}}{j}left ( frac{1}{j^{2}}H_{j}-frac{1}{j}frac{pi ^{2}}{6}+frac{1}{j}H_{j}^{left ( 2
ight )}
ight ) \
&=frac{pi ^{3}}{24}ln^{2}2+ln 2frac{pi }{16}zeta left ( 3
ight ) \
&~~~+frac{pi }{8}sum_{j=1}^{infty }frac{left ( -1
ight )^{j}}{j}left ( frac{2}{j^{2}}H_{j}+frac{2}{j}H_{j}^{left ( 2
ight )}-frac{3}{j^{3}}
ight )-frac{pi ^{5}}{576} \
&=frac{pi ^{3}}{24}ln^{2}2+ln 2frac{pi }{16}zeta left ( 3
ight )+frac{11pi ^{5}}{5760}+frac{pi }{4}sum left ( -1
ight )^{j}left ( frac{1}{j^{3}}H_{j}+frac{1}{j^{2}}H_{j}^{left ( 2
ight )}
ight ) \
&=frac{pi ^{3}}{24}ln^{2}2+ln 2frac{pi }{16}zeta left ( 3
ight )-frac{pi ^{5}}{960}-frac{pi }{16}sum_{j=1}^{infty }frac{H_{2j}}{j^{3}}
end{align*}]
Using the known result
[sum_{n=1}^{infty }frac{H_{2n}}{n^{3}}=-frac{pi ^{4}}{15}-frac{1}{3}pi ^{2}ln^{2}2+frac{ln^{4}2}{3}+8mathrm{Li}_{4}left ( frac{1}{2}
ight )+7ln 2zeta left ( 3
ight )
]
So here is the final result:
[Largeoxed{displaystyle egin{align*}
int_{0}^{frac{pi }{2}}x^{2}lnleft ( sin x
ight )lnleft ( cos x
ight )mathrm{d}x&=color{blue}{frac{pi ^{3}}{16}ln^{2}2+frac{pi ^{5}}{320}-frac{3}{8}ln 2zeta left ( 3
ight )}\
&~~~color{blue}{-frac{pi }{48}ln^{4}2-frac{1}{2}mathrm{Li}_{4}left ( frac{1}{2}
ight )}
end{align*}}]