Vrftsjtryk

It is not hard to find that $int_0^frac{pi}{2}ln{(sin{x})}dx=-frac{pi}{2}ln2$.

https://socratic.org/questions/how-do-you-prove-that-the-integral-of-ln-sin-x-on-the-interval-0-pi-2-is-converg

Moreover wolfram can compute that
$$int_0^frac{pi}{2}ln^2{(sin{x})}dx=frac{pi^3+3piln^24}{24}$$
$$int_0^frac{pi}{2}ln^3{(sin{x})}dx=-frac{pi}{16}(12zeta(3)+ln^34+pi^2ln4)$$
and $$int_0^frac{pi}{2}ln^4{(sin{x})}dx=pizeta(3)ln{8}+frac{19pi^5}{480}+frac{pi}{2}ln^42+frac{pi^3}{4}ln^22$$

How to prove these formulas?
In general, How to evaluate the integral $displaystyleint_0^frac{pi}{2}ln^n{(sin{x})}dx$ ?

Thanks in advance.

$begin{align}J_n=int_0^{frac{pi}{2}}ln^n(sin x),dxend{align}$

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

$begin{align}J_n=int_0^1 frac{ln^n(x)}{sqrt{1-x^2}},dxend{align}$

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

$begin{align}J_n=frac{1}{2^{n+1}}int_0^1 frac{x^{-frac{1}{2}}ln^n(x)}{sqrt{1-x}},dxend{align}$

Consider,

$begin{align}B_n(s)&=frac{1}{2^{n+1}}int_0^1 x^{-frac{1}{2}+s}(1-x)^{-frac{1}{2}},dx\
&=frac{1}{2^{n+1}}text{B}left(frac{1}{2}+s,frac{1}{2}right)\
&=frac{1}{2^{n+1}}frac{Gammaleft(frac{1}{2}+sright)Gammaleft(frac{1}{2}right)}{Gamma(1+s)}\
&=frac{sqrt{pi}}{2^{n+1}}frac{Gammaleft(frac{1}{2}+sright)}{Gamma(1+s)}
end{align}$

$begin{align}boxed{J_n=frac{partial^n B_n(s)}{partial s^n}big|_{s=0}}end{align}$

NB:
$text{B}(,)$ denotes the Beta Euler function.

Example:

Let $n=1$.

$begin{align}J_1=frac{sqrt{pi}}{4}frac{Gamma^primeleft(frac{1}{2}right)Gamma(1)-Gammaleft(frac{1}{2}right)Gamma^prime(1)}{Gamma^2(1)}end{align}$

But,

$begin{align}Gamma^primeleft(frac{1}{2}right)=-(gamma+2ln 2)sqrt{pi}end{align}$

$begin{align}Gammaleft(frac{1}{2}right)=sqrt{pi}end{align}$

$begin{align}Gammaleft(1right)=1end{align}$

$begin{align}Gamma^primeleft(1right)=-gammaend{align}$

Therefore,

$begin{align}J_1&=frac{sqrt{pi}}{4}times frac{-(gamma+2ln 2)sqrt{pi}+gammasqrt{pi}}{1}\
&=boxed{-frac{1}{2}piln 2}
end{align}$

Addendum:

To compute $J_n$ one needs to be able to evaluate $Gamma^{(k)}(1),Gamma^{(k)}left(frac{1}{2}right)$ for $0leq kleq n$.

Fact 1):

$begin{align} Gamma^{(m+1)}(z)=sum_{k=0}^{m} binom{m}{k}Gamma^{(k)}(z)psi^{(m-k)}(z)end{align}$

(following from the definition of Digamma function and the use of Leibniz-Newton formula)

Fact 2):

For $0<|z-1|<1$,

$begin{align}psi(z)=-gamma+sum_{k=1}^{infty}(-1)^{k+1}zeta(k+1)(z-1)^kend{align}$

Therefore,

For $kgeq 1$,

$displaystyle psi^{(k)}(1)=(-1)^{k+1}k!zeta(k+1)$

Fact 3):

For $0<Re(z)<1$,

$displaystyle psi(z)+psi(1-z)=picot(pi x)$

(allowing to compute $displaystyle psi^{(k)}left(frac{1}{2}right)$)

It's

$displaystyleintlimits_0^{pi/2} (sin x)^z dx = {binom{frac{z}{2}}{frac{1}{2}}}^{-1} = 2^{-z} sinfrac{pi z}{2} sumlimits_{k=0}^infty {binom z k} frac{(-1)^k}{z-2k}$

with $enspacedisplaystyle (sin x)^z = sumlimits_{n=0}^infty z^n frac{(ln sin x)^n}{n!}enspace$ , $enspace displaystyle 2^{-z} = sumlimits_{n=0}^infty z^n frac{(-ln 2)^n}{n!}enspace$ ,

$displaystyle frac{1}{z}sinfrac{pi z}{2} = sumlimits_{,,n=0\n,text{even}}^infty z^n frac{(-1)^{n/2}(frac{pi}{2})^{n+1}}{(n+1)!}$

and $enspaceenspacedisplaystyle sumlimits_{n=0}^infty z^n a_n := zsumlimits_{k=0}^infty {binom z k} frac{(-1)^k}{z-2k}enspace$ , $enspace$ where $,a_n,$ is calculated below.

It follows by coefficient comparison:

$$ frac{d^n}{dz^n}{binom{frac{z}{2}}{frac{1}{2}}}^{-1}|_{z=0}= intlimits_0^{pi/2} (ln sin x)^n dx = n! sumlimits_{k=0}^n a_{n-k} sumlimits_{,,j=0 \ j,text{even}}^k frac{(-ln 2)^{k-j}}{(k-j)!} frac{(-1)^{j/2}(frac{pi}{2})^{j+1}}{(j+1)!}$$

$a_0=1$ , $,,a_1=0$

Using the definitions

$displaystyle sumlimits_{k=0}^n left[ begin{array}{c} n \ k end{array} right] x^k := x(x+1)…(x+n-1) enspaceenspace$ (see Stirling numbers of the first kind) $enspace$ and

$displaystyle zeta_n(m):=sumlimits_{k=1}^infty frac{1}{k^m}left(frac{n!}{(k-1)!}left[begin{array}{c} k \ n+1 end{array} right]right)enspace $
(as in Evaluate $int_{0}^{pi }theta ^{3}log^{3}left ( 2sinfrac{theta }{2} right )mathrm{d}theta $)

we get for $,kgeq 2,$ the coefficients $enspacedisplaystyle a_k=sumlimits_{j=1}^{k-1}frac{(-1)^{j-1}}{2^{k-j}}frac{zeta_{j-1}(k-j+1)}{(j-1)!}enspace$ .

Note:

$displaystyle frac{zeta_{v}(n-v+2)}{v!}=frac{zeta_{n-v}(v+2)}{(n-v)!}enspaceenspace$ , $enspaceenspace$ special case $n:=v,$: $enspacedisplaystyle frac{zeta_{v}(2)}{v!}=zeta(v+2)$

$displaystyle frac{zeta_v(3)}{v!}= zeta_1(v+2)=(1+frac{v}{2})zeta(v+3) - frac{1}{2} sumlimits_{j=1}^v zeta(j+1)zeta(v+2-j)$

...

In here, part Expansion by harmonic numbers, is $enspacedisplaystyle w(n,m):=frac{m!}{(n-1)!}left[ begin{array}{c} n \ {m+1} end{array} right]enspace$ with

the recursion $enspacedisplaystyle w(n,m)= delta_{m,0} + sumlimits_{j=0}^{m-1}(-1)^j frac{(m-2+j)!}{(m-2)!}H_{n-1}^{(j+1)} w(n,m-1-j) enspace$ and

the $,m$-order harmonic number $,H_n^{(m)},$ so that we can also write $enspacedisplaystyle zeta_n(m):=sumlimits_{k=1}^infty frac{w(k,n)}{k^m},$ .