Logarithmic-Trigonometric积分系列（二）
[Largedisplaystyle int_0^{pi/2}ln^2(sin x)ln(cos x) an x \,{
m d}x
]
(Largemathbf{Solution:})
Let (J) donates the integral and it is easy to see that
[egin{align*}
J&=int_0^{pi/4}ln^2(sin x)ln(cos x) an x \,{
m d}x+
int_{pi/4}^{pi/2}ln^2(sin x)ln(cos x) an x \,{
m d}xcr
&=int_0^{pi/4}ln^2(sin x)ln(cos x) an x \,{
m d}x+
int_{0}^{pi/4}ln^2(cos x)ln(sin x)cot x \,{
m d}xcr
end{align*}]
Now, to calculate (J) we make the substitution (tleftarrowsin^2x):
[J=frac{1}{16}int_0^1frac{ln(1-u)}{1-u}ln^2(u)\,{
m d}u
]
But
[frac{ln(1-u)}{1-u}=-left(sum_{n=0}^infty u^n
ight)left(sum_{n=1}^infty frac{u^n}{n}
ight)
=-sum_{n=1}^infty H_nu^n]
where (H_n=displaystylesum_{k=1}^n frac{1}{k}).Hence
[J=-frac{1}{16}sum_{n=1}^infty H_nint_0^1u^nln^2(u){
m d}u
=-frac{1}{8}sum_{n=1}^inftyfrac{ H_n}{(n+1)^3}]
The sum (displaystylesum_{n=1}^inftyfrac{H_n}{n^3}) is known, it can be evaluated as follows, first we have
[H_n=sum_{k=1}^inftyleft(frac{1}{k}-frac{1}{k+n}
ight)=
sum_{k=1}^infty frac{n}{k(k+n)}]
Thus
[sum_{n=1}^inftyfrac{H_n}{n^3}=sum_{k,ngeq1}frac{1}{n^2k(n+k)}
=sum_{k,ngeq1}frac{1}{k^2n(n+k)}]
Taking the half sum we find
[sum_{n=1}^inftyfrac{H_n}{n^3}=frac{1}{2}sum_{k,ngeq1}frac{1}{kn(k+n)}left(frac{1}{k}+frac{1}{n}
ight)=
frac{1}{2}sum_{k,ngeq1}frac{1}{k^2n^2}=frac{1}{2}zeta^2(2)]
then we obtain
[Largeoxed{displaystyle egin{align*}
int_0^{pi/2}ln^2(sin x)ln(cos x) an x \,{
m d}x&=frac{1}{8}zeta(4)-frac{1}{16}zeta^2(2)\
&=color{blue}{-frac{pi^4}{2880}}
end{align*}}]