Ytukyg

Show that $displaystyle{int_0^1 log log left(frac{1}{x}right) frac{dx}{1+x^2} = frac{pi}{2}log left(sqrt{2pi} Gammaleft(frac{3}{4}right) / Gammaleft(frac{1}{4}right)right)}$

This question was posted as part of this question:

Solve the integral $S_k = (-1)^k int_0^1 (log(sin pi x))^k dx$

I cannot think of a change of variable nor other integrating methods. May be there is a known method that I am missing.

By the substitution $x = e^{-t}$, we find that

$$begin{align*}
int_{0}^{1} frac{(log (1/x))^s}{1+x^2} ; dx
&= int_{0}^{infty} frac{t^s e^{-t}}{1 + e^{-2t}} ; dt \
&= int_{0}^{infty} sum_{n=0}^{infty} (-1)^n t^s e^{-(2n+1)t} ; dt \
&= sum_{n=0}^{infty} (-1)^n , frac{Gamma(s+1)}{(2n+1)^{s+1}} \
&= Gamma(s+1)L(s+1, chi_4),
end{align*}$$

where $L(s, chi_4)$ is the Dirichlet L-function of the unique non-principal character $chi_4$ to the modulus 4. Often it is denoted as $beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula

$$int_{0}^{1} frac{log log (1/x)}{1+x^2} ; dx = psi_0(1) beta(1) + beta'(1),$$

and the problem reduces to find the value of $beta(1)$ and $beta'(1)$. Note that $beta(1) = frac{pi}{4}$ is just the Gregory series. For $beta'(1)$, we first notice that the following functional equation holds.

$$ beta(s)=left(frac{pi}{2}right)^{s-1} Gamma(1-s) cos left( frac{pi s}{2} right),beta(1-s). $$

This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $beta'(0)$. For $0 < s$, we have

$$begin{align*}
-beta'(s)
&= sum_{n=1}^{infty} left[ frac{log(4n+1)}{(4n+1)^s} - frac{log(4n-1)}{(4n-1)^s} right] \
&= sum_{n=1}^{infty} frac{1}{(4n)^s} left[ log left( frac{4n+1}{4n-1} right) - frac{1}{2n} right] + 2^{-2s-1}zeta(s+1) \
& qquad + sum_{n=1}^{infty} left( frac{1}{(4n+1)^s} - frac{1}{(4n)^s} right) log (4n+1) \
& qquad + sum_{n=1}^{infty} left( frac{1}{(4n)^s} - frac{1}{(4n-1)^s} right) log (4n-1) \
& =: A(s) + 2^{-2s-1}zeta(s+1) + B(s) + C(s).
end{align*}$$

We first estimate $B(s)$. As $n to infty$, we have

$$ log left( frac{4n}{4n+1} right) = -frac{1}{4n} + Oleft( frac{1}{n^2} right), quad log left( frac{4n}{4n-1} right) = frac{1}{4n} + Oleft( frac{1}{n^2} right). $$

Thus when $s to 0$,

$$begin{align*}
B(s)
&= sum_{n=1}^{infty} frac{1}{(4n)^s} left[ expleft( s log left( frac{4n}{4n+1} right) right) - 1 right] left[ log (4n) - log left(frac{4n}{4n+1} right) right] \
&= sum_{n=1}^{infty} frac{1}{(4n)^s} left[ - frac{s}{4n} + O left(frac{s^2}{n^2} right) right] left[ log (4n) + O left(frac{1}{n} right) right] \
&= -s 2^{-2s-2} sum_{n=1}^{infty} frac{1}{n^{s+1}} log (4n) + O(s) \
&= s 2^{-2s-2} left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).
end{align*}$$

Similar consideration also shows that

$$ C(s) = s 2^{-2s-2} left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).$$

Thus we have

$$ 2^{-2s-1}zeta(s+1) + B(s) + C(s) = 2^{-2s-1} left[ zeta(s+1) + s zeta'(s+1) - s zeta(s+1) log 4 right] + O(s). $$

But since

$$zeta(1+s) = frac{1}{s} + gamma + O(s),$$

we have

$$ lim_{sdownarrow 0} left( 2^{-2s-1}zeta(s+1) + B(s) + C(s) right) = frac{gamma}{2} - log 2.$$

For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that

$$ lim_{sdownarrow 0} A(s) = sum_{n=1}^{infty} left[ log left( frac{4n+1}{4n-1} right) - frac{1}{2n} right]. $$

Let $L$ denote this limit. Then by Stirling's formula,

$$begin{align*}
e^{L}
& stackrel{Ntoinfty}{sim} prod_{n=1}^{N} left( frac{4n+1}{4n-1} right) e^{-1/2n}
sim frac{e^{-gamma/2}}{sqrt{N}} prod_{n=1}^{N} left( frac{n+(1/4)}{n-(1/4)} right) \
& sim frac{e^{-gamma/2}}{sqrt{N}} frac{Gammaleft(frac{3}{4}right)}{Gammaleft(frac{5}{4}right)} frac{Gammaleft(N+frac{5}{4}right)}{Gammaleft(N+frac{3}{4}right)}
sim frac{e^{-gamma/2}}{sqrt{N}} frac{Gammaleft(frac{3}{4}right)}{Gammaleft(frac{5}{4}right)} frac{left( frac{N + (5/4)}{e} right)^{N+frac{5}{4}}}{left( frac{N + (3/4)}{e} right)^{N+frac{3}{4}}} \
& sim e^{-gamma/2} frac{Gammaleft(frac{3}{4}right)}{Gammaleft(frac{5}{4}right)}
= 4 e^{-gamma/2} frac{pi sqrt{2}}{Gammaleft(frac{1}{4}right)^2},
end{align*}$$

where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain

$$-beta'(0) = log (2 pi sqrt{2}) - 2 log Gammaleft(frac{1}{4}right) .$$

Now taking logarithmic differntiation to the functional equation, we have

$$ frac{beta'(s)}{beta(s)} = logleft(frac{pi}{2}right) - psi_0 (1-s) - frac{pi}{2} tan left( frac{pi s}{2} right) - frac{beta'(1-s)}{beta(1-s)}. $$

Taking $s = 0$, we have

$$ frac{beta'(0)}{beta(0)} = logleft(frac{pi}{2}right) + gamma - frac{beta'(1)}{beta(1)} quad Longrightarrow quad beta'(1) = beta(1) left[ logleft(frac{pi}{2}right) + gamma - frac{beta'(0)}{beta(0)} right]. $$

But again by the functional equation, we have $beta(0) = frac{1}{2}$. Therefore

$$ beta'(1) = frac{pi}{4} left[ gamma + 2 log 2 + 3 log pi - 4 log Gammaleft(frac{1}{4}right) right], $$

and hence

$$ int_{0}^{1} frac{log log (1/x)}{1+x^2} ; dx = frac{pi}{4} left[ 2 log 2 + 3 log pi - 4 log Gammaleft(frac{1}{4}right) right], $$

which is identical to the proposed answer.

$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle #1 rightrangle}
newcommand{braces}[1]{leftlbrace #1 rightrbrace}
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}
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newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
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$ds{int_{0}^{1}lnpars{lnpars{1 over x}},{dd x over 1 + x^{2}}
={pi over 2},lnpars{root{vphantom{large A}2pi},
{Gammapars{3/4} over Gammapars{1/4}}}: {Large ?}}$

With $ds{x to 1/x}$:
$$
color{#f00}{int_{0}^{1}lnpars{lnpars{1 over x}},{dd x over 1 + x^{2}}}=
int_{infty}^{1}lnpars{lnpars{x}},pars{-,{dd x/x^{2} over 1 + 1/x^{2}}}
=int_{1}^{infty}{lnpars{lnpars{x}} over 1 + x^{2}},dd x
$$

With $x equiv expo{t}quadiffquad t = lnpars{x}$:
begin{align}
&color{#f00}{int_{0}^{1}lnpars{lnpars{1 over x}},{dd x over 1 + x^{2}}}=
int_{0}^{infty}{lnpars{t} over 1 + expo{2t}},expo{t}dd t
=int_{0}^{infty}lnpars{t}expo{-t},{1 over 1 + expo{-2t}},dd t
\[3mm]&=int_{0}^{infty}lnpars{t}expo{-t},
sum_{ell = 0}^{infty}pars{-1}^{ell}expo{-2ell t},dd t
=sum_{ell = 0}^{infty}pars{-1}^{ell}lim_{mu to 0}partiald{}{mu}bracks{%
int_{0}^{infty}t^{mu}expo{-pars{2ell + 1}t},dd t}
\[3mm]&=sum_{ell = 0}^{infty}pars{-1}^{ell}lim_{mu to 0}partiald{}{mu}bracks{%
{1 over pars{2ell + 1}^{mu + 1}}
overbrace{int_{0}^{infty}t^{mu}expo{-t},dd t}^{ds{Gammapars{mu + 1}}} }
end{align}
where $Gammapars{z}$ is the
Gamma Function.

begin{align}
&color{#f00}{int_{0}^{1}lnpars{lnpars{1 over x}},{dd x over 1 + x^{2}}}=
sum_{ell = 0}^{infty}pars{-1}^{ell}lim_{mu to 0}partiald{}{mu}bracks{%
{Gammapars{mu + 1} over pars{2ell + 1}^{mu + 1}}}
\[3mm]&=
lim_{mu to 0}partiald{}{mu}bracks{%
Gammapars{mu + 1}sum_{ell = 0}^{infty}
{pars{-1}^{ell}over pars{2ell + 1}^{mu + 1}}}
\[3mm]&=
lim_{mu to 0}braces{%
Gamma'pars{mu + 1}sum_{ell = 0}^{infty}
{pars{-1}^{ell}over pars{2ell + 1}^{mu + 1}}
+Gammapars{mu + 1}partiald{}{mu}bracks{%
sum_{ell = 0}^{infty}{pars{-1}^{ell}over pars{2ell + 1}^{mu + 1}}}}
\[3mm]&=-gammasum_{ell = 0}^{infty}{pars{-1}^{ell}over 2ell + 1}
+
lim_{mu to 0}partiald{}{mu}
sum_{ell = 0}^{infty}{pars{-1}^{ell}over pars{2ell + 1}^{mu + 1}}tag{1}
end{align}
In this result, we used $Psipars{1} = -gamma$ and $Gammapars{1} = 1$ where
$Psipars{z} equiv ddlnGammapars{z}/dd z$ is the
Digamma Function and $gamma$ is the
Euler-Mascheroni constant.

The first $ell$-sum in the right member of $pars{1}$ is given by:
begin{align}
&sum_{ell = 0}{pars{-}^{ell} over 2ell + 1}=
sum_{ell = 0}pars{{1 over 4ell + 1} - {1 over 4ell + 3}}
={1 over 8}sum_{ell = 0}{1 over pars{ell + 1/4}pars{ell + 3/4}}
\[3mm]&=-,{1 over 4}bracks{Psipars{1 over 4} - Psipars{3 over 4}}
= {pi over 4}
end{align}
where we used the identities:
begin{align}
sum_{ell = 0}^{infty}{1 over pars{ell + z_{0}}pars{ell + z_{1}}}
&={Psipars{z_{0}} - Psipars{z_{1}} over z_{0} - z_{1}}tag{1.1}
\[3mm]Psipars{z} - Psipars{1 - z} &= -picotpars{pi z}tag{1.2}
end{align}
$$
mbox{Then,}quad
color{#f00}{int_{0}^{1}lnpars{lnpars{1 over x}},{dd x over 1 + x^{2}}}=
-,{1 over 4},gammapi + lim_{mu to 0}partiald{}{mu}
sum_{ell = 0}^{infty}{pars{-1}^{ell}over pars{2ell + 1}^{mu + 1}}tag{2}
$$
Also,
begin{align}
sum_{ell = 0}^{infty}{pars{-1}^{ell}over pars{2ell + 1}^{mu + 1}}
&=sum_{ell = 0}^{infty}{1 over bracks{2pars{2ell} + 1}^{mu + 1}}
-sum_{ell = 0}^{infty}{1 over bracks{2pars{2ell + 1} + 1}^{mu + 1}}
\[3mm]&=2^{-2mu - 2}bracks{%
sum_{ell = 0}^{infty}{1 over pars{ell + 1/4}^{mu + 1}}
-sum_{ell = 0}^{infty}{1 over pars{ell + 3/4}^{mu + 1}}}
\[3mm]&=2^{-2mu - 2}bracks{%
zetapars{mu + 1,{1 over 4}} - zetapars{mu + 1,{3 over 4}}}
end{align}
where $ds{zetapars{s,q} equiv sum_{n = 0}^{infty}{1 over pars{q + n}^{s}}}$
with $Repars{s} > 1$ and $Repars{q} > 0$ i s the
Hurwitz Zeta Function or/and Generalizated Zeta Funcion .

So,
begin{align}
&lim_{mu to 0}partiald{}{mu}sum_{ell = 0}^{infty}{pars{-1}^{ell}over pars{2ell + 1}^{mu + 1}}
\[3mm]&=-,{1 over 4},lnpars{2}
underbrace{overbrace{sum_{ell = 0}^{infty}{1 over pars{ell + 3/4}pars{ell + 1/4}}}
^{ds{2bracks{Psipars{3/4} - Psipars{1/4}} = 2pi}}}
_{ds{mbox{See} pars{1.1} mbox{and} pars{1.2}}}
+ {1 over 4}
overbrace{partiald{}{mu}bracks{%
zetapars{mu,{1 over 4}} - zetapars{mu,{3 over 4}}}_{mu = 1}}
^{ds{-gamma_{1}pars{1/4} + gamma_{1}pars{3/4}}}
end{align}
where $gamma_{n}pars{z}$ is a
Generalizated Stieltjes Constant
.

With this result, $pars{2}$ is reduced to:
begin{align}
color{#f00}{int_{0}^{1}lnpars{lnpars{1 over x}},{dd x over 1 + x^{2}}}
&=-,{1 over 4},braces{%
pibracks{%
gamma + 2lnpars{2}} + gamma_{1}pars{1 over 4} - gamma_{1}pars{3 over 4}}
tag{3}
end{align}
The difference $gamma_{1}pars{1/4} - gamma_{1}pars{3/4}$ is evaluated with the 1846 Carl Malmsten identity
:
$$
gamma_{1}pars{m over n} - gamma_{1}pars{1 - {m over n}}
=-pibracks{gamma + lnpars{2pi n}}cotpars{mpi over n}
+ 2pisum_{ell = 1}^{n - 1}
sinpars{{2pi m over n},ell}lnpars{Gammapars{ell over n}}
$$

With $m = 1$ and $n = 4$:
begin{align}
&gamma_{1}pars{1 over 4} - gamma_{1}pars{3 over 4}
\[3mm]&=-pibracks{gamma + lnpars{8pi}}cotpars{pi over 4}
+ 2pisum_{ell = 1}^{3}sinpars{pi,ell over 2}
lnpars{Gammapars{ell over 4}}
\[3mm]&=-pibracks{gamma + 2lnpars{2} + lnpars{2pi}}
\[3mm]&phantom{=}
+ 2pibracks{sinpars{pi over 2}lnpars{Gammapars{1 over 4}}
+ sinpars{pi}lnpars{Gammapars{1 over 2}}
+ sinpars{3pi over 2}Gammapars{3 over 4}}
\[3mm]&=-pibracks{gamma + 2lnpars{2} + lnpars{2pi}}
+2pibracks{lnpars{Gammapars{1 over 4}} - lnpars{Gammapars{3 over 4}}}
\[3mm]&=-pibracks{gamma + 2lnpars{2}} - 2pibracks{%
lnpars{root{2pi}} +lnpars{Gammapars{3 over 4} over Gammapars{1 over 4}}}
\[3mm]&=-pibracks{gamma + 2lnpars{2}}
-2pilnpars{root{2pi},{Gammapars{3 over 4} over Gammapars{1 over 4}}}
end{align}

By replacing this result in $pars{3}$, we find:
$$color{#00f}{large%
int_{0}^{1}lnpars{lnpars{1 over x}},{dd x over 1 + x^{2}}
={pi over 2},lnpars{root{vphantom{large A}2pi},
{Gammapars{3/4} over Gammapars{1/4}}}}
$$

As an 'extra-bonus' we can use the identity
$ds{Gammapars{z} = {pi over Gammapars{1 - z}sinpars{pi z}}}$ to 'kill' one of the $Gamma,$'s functions:
$ds{Gammapars{1 over 4} = {root{2}pi over Gammapars{3/4}}}$ which yields:
$$
int_{0}^{1}lnpars{lnpars{1 over x}},{dd x over 1 + x^{2}}
={pi over 2},lnpars{Gamma^{,2}pars{3/4} over root{pi}}
$$

See also V. Adamchik's formula $$int_0^1 frac{x^{p-1}}{1+x^n}log log frac{1}{x}dx = frac{gamma+log(2n)}{2n}(psi(frac{p}{2n})-psi(frac{n+p}{2n}))+frac{1}{2n}(zeta'(1,frac{p}{2n})-zeta'(1,frac{n+p}{2n}))$$ in http://dx.doi.org/10.1145/258726.258736 .

The above integral was long-time known as the Vardi's integral. Quite recently, an interesting research work devoted to this integral was published by Iaroslav Blagouchine. It appears that this integral was first evaluated by Carl Malmsten in 1842 (and not by Ilan Vardi in 1988). Blagouchine describes two different methods for its evaluation: the original Malmsten's method and the contour integration method. In the same paper, numerous integrals similar to the above one are also treated, and it is shown that many of them may be evaluated by the contour integration methods.

More recent paper by Blagouchine (mentioned above in the context of Malmsten's formula derived in 1846) is also integersting, but it is less directly related to the integral in question than the first one.