[Largedisplaystyle int_0^{1} frac{arccos^4 left(x^2
ight)}{sqrt{1-x^2}}\,mathrm{d}x
]
(Largemathbf{Solution:})
Let (I) denote the integral. Using the substitution (x=sqrt{cos t}), we get
[I=frac{1}{2}int_0^{frac{pi}{2}}frac{t^4sin(t)}{sqrt{cos t-cos^2t}}mathrm{d}t=frac{1}{sqrt{2}}int_0^{frac{pi}{2}}frac{t^4cosleft(dfrac{t}{2}
ight)}{sqrt{1-2sin^2left(dfrac{t}{2}
ight)}}mathrm{d}t
]
Next, substitute (displaystyle 2sin^2left(frac{t}{2}
ight)=sin^2 heta) to get
[I=16int_0^{frac{pi}{2}}left[sin^{-1}left(frac{sin heta}{sqrt{2}}
ight)
ight]^4; mathrm{d} heta
]
Now, plug in the series (displaystyle (sin^{-1}z)^4=sum_{k=1}^infty frac{H_{k-1}^{(2)}(2z)^{2k}}{k^2dbinom{2k}{k}}) and integrate termwise:
[egin{align*} I &= 24int_0^{frac{pi}{2}}left(sum_{k=1}^infty frac{H_{k-1}^{(2)}2^k}{k^2dbinom{2k}{k}}sin^{2k}( heta)
ight); mathrm{d} heta= 24sum_{k=1}^infty frac{H_{k-1}^{(2)}2^k}{k^2dbinom{2k}{k}}int_0^{frac{pi}{2}}sin^{2k}( heta)mathrm{d} heta \ &= 12pi sum_{k=1}^infty frac{H_{k-1}^{(2)}2^k}{k^2dbinom{2k}{k}}left(frac{dbinom{2k}{k}}{2^{2k}}
ight)= 12pisum_{k=1}^infty frac{H_{k-1}^{(2)}}{k^2 2^k}= 12pisum_{k=1}^infty frac{zeta(2)-psi_1(k)}{k^2 2^k} \ &= 12pizeta(2) ext{Li}_2left(frac{1}{2}
ight)-12pisum_{n=1}^inftyfrac{psi_1(k)}{k^2 2^k} end{align*}
]
For the last series
[egin{align*} sum_{n=1}^inftyfrac{psi_1(n)}{2^n n^2} &= -sum_{n=1}^inftypsi_1(n)left(frac{ln(2)}{2^n n}+int_0^{frac{1}{2}} x^{n-1}ln(x)mathrm{d}x
ight) \ &= -ln(2)sum_{n=1}^inftyfrac{psi_1(n)}{2^n n}-int_0^{1over 2}frac{ln(x)}{1-x}left(zeta(2)- ext{Li}_2(x)
ight)mathrm{d}x ag{1} end{align*}
]
1.Evaluation of(displaystyle int_0^{1over 2}frac{ln(x)}{1-x}left(zeta(2)- ext{Li}_2(x)
ight)mathrm{d}x)
Using the identity ( ext{Li}_2(x)+ ext{Li}_2(1-x)=zeta(2)-ln(x)ln(1-x)), we have
[egin{align*} int_0^{1over 2}frac{ln(x)}{1-x}left(zeta(2)- ext{Li}_2(x)
ight)mathrm{d}x &= color{red}{int_0^{1over 2}frac{ln^2(x)ln(1-x)}{1-x}mathrm{d}x}+int_0^{1over 2}frac{ln(x) ext{Li}_2(1-x)}{1-x}mathrm{d}x \ &={-frac{pi^4}{360}-frac{ln^4(2)}{4}}+{left(frac{1}{2} ext{Li}_2^2(1-x)
ight)Bigg|_0^{1over 2}} \ &= -frac{19pi^4}{1440}-frac{pi^2}{24}ln^2(2)-frac{ln^4(2)}{8} ag{2} end{align*}
]
The red integral was evaluated using equation 12 on page 310 of Leonard Lewin - Polylogarithms and associated functions. The proof is discussed on page 203 and 204.
2.Evaluation of(displaystyle sum_{n=1}^inftyfrac{psi_1(n)}{2^n n})
[egin{align*} sum_{n=1}^inftyfrac{psi_1(n)}{2^n n}& = sum_{n=1}^infty psi_1(n)int_0^{1over 2}x^{n-1}mathrm{d}x \
&= int_0^{frac{1}{2}}frac{zeta(2)- ext{Li}_2(x)}{1-x}mathrm{d}x\
&= color{Purple}{int_0^{1over 2}frac{ln(x)ln(1-x)}{1-x}mathrm{d}x}+int_0^{1over 2}frac{ ext{Li}_2(1-x)}{1-x}mathrm{d}x\
&= {frac{ln^3(2)}{2}+frac{1}{2}int_0^{frac{1}{2}}frac{ln^2(1-x)}{x}mathrm{d}x }+{zeta(3)- ext{Li}_3left(frac{1}{2}
ight)} \ &= {frac{ln^3(2)}{2}+frac{1}{2}left( -ln^3(2)-2ln(2) ext{Li}_2left(frac{1}{2}
ight)+2zeta(3)-2 ext{Li}_3left( frac{1}{2}
ight)
ight) }\
&~~~+{zeta(3)- ext{Li}_3left(frac{1}{2}
ight)} \ &= frac{pi^2}{12}ln(2)+frac{ln^3(2)}{6}+frac{zeta(3)}{4} ag{3} end{align*}]
The purple integral was evaluated using the generalized result found in this thread.
3.The Final Answer
Substitute (2) and (3) into (1) to get
[largeoxed{displaystyle sum_{n=1}^inftyfrac{psi_1(n)}{2^n n^2}=color{Teal}{frac{19pi^4}{1440}-frac{pi^2}{24}ln^2(2)-frac{ln^4(2)}{24}-frac{zeta(3) ln(2)}{4}}}
]
Hence The final result of the initial integral follows from the above answer
[Largeoxed{displaystyle int_0^{1} frac{arccos^4 left(x^2
ight)}{sqrt{1-x^2}}\,mathrm{d}x=color{Blue}{ frac{pi^5}{120}+frac{pi }{2}ln^4 2 + 3pi zeta(3)ln 2 - frac{pi^{3} }{2}ln^2 2}}
]