Ytukyg

I saw in a paper by @Jack D'aurizio the following integral
$$I=int_0^{pi/2}log^2(cos^2x)mathrm{d}x=frac{pi^3}6+2pilog^2(2)$$
Below is my attempt.

$$I=4int_0^{pi/2}log^2(cos x)mathrm{d}x$$
Then we define
$$F(a)=int_0^{pi/2}log^2(acos x)mathrm{d}x$$
So we have
$$F'(a)=frac2aint_0^{pi/2}log(acos x)mathrm{d}x$$
Which I do not know how to compute. How do I proceed? Thanks.

Let $$I(a)=int_0^{frac {pi}{2}} (cos^2 x)^a dx$$

Hence we need $I''(0)$.

Now recalling the definition of Beta function we get $$I(a)=frac 12 Bleft(a+frac 12 ,frac 12right)=frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}$$

Hence we have $$I''(a) =frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}left(left[psi^{(0)}left(a+frac 12 right)-psi^{(0)}(a+1)right]^2 +psi^{(1)}left(a+frac 12 right)-psi^{(1)}(a+1)right) $$

Substituting $a=0$ in above formula yields the answer.

begin{align}J=int_0^{frac{pi}{2}} ln^2left(cos xright),dxend{align}

Observe that,

begin{align}I&=4J\
J&=int_0^{frac{pi}{2}} ln^2left(sin xright),dx\
int_0^{frac{pi}{2}} lnleft(sin xright),dx&=int_0^{frac{pi}{2}} lnleft(cos xright),dx
end{align}

(change of variable $y=dfrac{pi}{2}-x$ )

begin{align}
K&=int_0^{frac{pi}{2}} ln^2 left(2sin xcos xright),dx\
&=int_0^{frac{pi}{2}} ln^2 left(sinleft(2xright)right),dx\
end{align}

Perform the change of variable $y=2x$,

begin{align}
K&=frac{1}{2}int_0^{pi} ln^2 left(sin xright),dx\
&=frac{1}{2}int_0^{frac{pi}{2}} ln^2 left(sin xright),dx+frac{1}{2}int_{frac{pi}{2}}^pi ln^2 left(sin xright),dx\
end{align}

In the latter integral perform the change of variable $y=dfrac{pi}{2}-x$ and recall $sinleft(pi-xright)=sin x$ for $x$ real,

begin{align}
K&=int_0^{frac{pi}{2}} ln^2 left(sin xright),dx\
&=int_0^{frac{pi}{2}} ln^2 left(cos xright),dx\
&=J
end{align}

On the other hand,

begin{align}
K&=int_0^{frac{pi}{2}}left(ln 2+ln(sin x)+ln(cos x)right)^2 ,dx\
&=frac{pi}{2}ln^2 2+int_0^{frac{pi}{2}}ln^2(sin x),dx+int_0^{frac{pi}{2}}ln^2(cos x),dx+2ln 2int_0^{frac{pi}{2}}ln(cos x),dx+\
&2ln 2int_0^{frac{pi}{2}}ln(sin x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
&=frac{pi}{2}ln^2 2+2J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
end{align}

begin{align}L&=int_0^infty frac{ln^2 x}{1+x^2},dxend{align}

Perform the change of variable $x=tan y$,

begin{align}L&=int_0^{frac{pi}{2}} ln^2left(tan xright),dx\
&=int_0^{frac{pi}{2}}left(lnleft(sin xright)-lnleft(cos xright)right)^2,dx\
&=2J-2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
end{align}

Therefore,

begin{align}K+L&=frac{pi}{2}ln^2 2+4J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx
end{align}

Therefore (recall $K=J$),

begin{align}J&=frac{1}{3}L-frac{pi}{6}ln^2 2-frac{4}{3}ln 2int_0^{frac{pi}{2}}ln(cos x),dxend{align}

On the other hand,

begin{align}L&=int_0^1 frac{ln^2 x}{1+x},dx+int_1^infty frac{ln^2 x}{1+x},dxend{align}

In the latter integral perform the change of variable $y=dfrac{1}{x}$,

begin{align}L&=2int_0^1 frac{ln^2 x}{1+x},dxend{align}

But it is well known that,

begin{align}int_0^1 frac{ln^2 x}{1+x^2},dx&=frac{pi^3}{16}\
int_0^{frac{pi}{2}}ln(cos x),dx&=-frac{1}{2}piln 2
end{align}

Therefore,

begin{align}J&=frac{pi^3}{24}-frac{pi}{6}ln^2 2-frac{4}{3}ln 2times -frac{1}{2}piln 2\
&=boxed{frac{pi^3}{24}+frac{1}{2}piln^2 2}\
end{align}

PS:
See: https://math.stackexchange.com/a/2942594/186817

(in this post i assume only the value of $zeta(4)$ )

Differentiation of the Beta function is a viable approach, but since we are dealing with a squared logarithm and not higher powers, it is faster to just exploit the Fourier (cosine) series of $log(cos^2theta)$ and Parseval's theorem. Given

$$-log(cos^2theta)=2log 2+2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta),tag{1} $$
since $int_{0}^{pi/2}cos(2jx)cos(2kx),dx =frac{pi}{4}delta(j,k)$, we immediately have
$$ int_{0}^{pi/2}log^2(cos^2theta),dtheta = 2pilog^2 2+pizeta(2).tag{2}$$

Since the LHS equals $int_{0}^{1}frac{4log^2 x}{sqrt{1-x^2}},dx$, we have just found the value of the hypergeometric series
$$ 8sum_{ngeq 0}frac{binom{2n}{n}}{4^n(2n+1)^3}=8cdotphantom{}_4 F_3left(tfrac{1}{2},tfrac{1}{2},tfrac{1}{2},tfrac{1}{2};tfrac{3}{2},tfrac{3}{2},tfrac{3}{2};1right)tag{3} $$
which also equals $int_{0}^{+infty}frac{log^2(1+t^2)}{1+t^2},dt$ or $frac{1}{2}int_{0}^{1}frac{log^2(x)}{sqrt{x(1-x)}},dx$.

Getting rid of the first roadblock:

Use the fact that $$log(acos x) = log a + logcos x$$
along with the result that
$$
int_0^{pi/2} logcos x,dx = -frac{pi}{2}log 2
$$
to get
$$
F'(a) = frac{pilog a}{a} -frac{pi}{2a}log 2
$$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

$ds{I equiv
int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x =
{pi^{3} over 6} + 2piln^{2}pars{2}: {LARGE ?}}$.

begin{align}
I & equiv
bbox[10px,#ffd]{int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x}
,,,stackrel{x mapsto pi/2 - x}{=},,,
4int_{0}^{pi/2}ln^{2}pars{sinpars{x}},dd x
\[5mm] & =
left.4,Reint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over ic z}
,rightvert_{ z = exppars{ic x}}
\[5mm] & =
left.4,Imint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over z}
,rightvert_{ z = exppars{ic x}}
\[1cm] & stackrel{mrm{as} epsilon to 0^{+}}{sim},,,
-, 4,
overbrace{Imint_{1}^{epsilon}ln^{2}
pars{1 + y^{2} over 2y},{dd y over y}}^{ds{= large 0}}
\[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,} -
4,Imint_{pi/2}^{0}bracks{lnpars{1 over 2epsilon} + pars{{pi over 2} - theta}ic}^{2},ic,ddtheta
\[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,,}
-4,Imint_{epsilon}^{1}bracks{lnpars{1 - x^{2} over 2x} + {pi over 2},ic}^{2}{dd x over x}
\[1cm] & =
4int_{0}^{pi/2}bracks{ln^{2}pars{2epsilon} -
pars{{pi over 2} - theta}^{2}},ddtheta
-4piint_{epsilon}^{1}lnpars{1 - x^{2} over 2x}
{dd x over x}
\[5mm] & stackrel{mrm{as} epsilon to 0^{+}}{sim}
bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
bracks{4piint_{0}^{1}{lnpars{1 - x^{2}} over x},dd x -
4piint_{epsilon}^{1}{lnpars{2x} over x},dd x}
\[5mm] & =
bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
bracks{2piint_{0}^{1}{lnpars{1 - x} over x},dd x +
2pilnpars{epsilon}lnpars{4epsilon}}
\[5mm] & stackrel{mrm{as} epsilon to 0}{to}
bbx{{pi^{3} over 6} + 2piln^{2}pars{2}} approx
8.1865
end{align}

Note that
$ds{int_{0}^{1}{lnpars{1 - x} over x},dd x =
-,mrm{Li}pars{1} = -,{pi^{2} over 6}}$.