Mathematics Asked by Iridescent on December 3, 2020
Gradshteyn&Ryzhik $3.554.5$ states that:
$$int_0^{infty } frac1x biggl( 2qe^{-x}-frac{sinh (q x)}{sinh left(frac{x}{2}right)} biggr) , dx=logbigl(cos (pi q) bigr)+2 log left(Gamma bigl(q+frac12 bigr)right)-log (pi )~, quad q^2<frac{1}{4}$$
It seems like to have sth to do with the Binet representation of log-gamma function, but I haven’t figured out how to solve it. Any kind of help will be appreciated.
The formula of the question follows easily from Ramanujan's generalization of Frullani's integral. See 'The Quarterly Reports of S. Ramanujan,' American Mathematical Monthly, Vol. 90, #8 Oct 1983, p 505-516. I won't give the conditions, but state it to establish the notation. Let $$ f(x)-f(infty)=sum_{k=0}^infty u(k)(-x)^k/k! quad,quad g(x)-g(infty)=sum_{k=0}^infty v(k)(-x)^k/k!$$ $$ f(0)=g(0) quad,quad f(infty)=g(infty) $$ Then $$int_0^infty frac{dx}{x} big(f(ax) - g(bx) big)= big(f(0)-f(infty) big)Big( log{(b/a)} + frac{d}{ds} log{Big(frac{v(s)}{u(s)}Big)}Big|_{s=0} Big) $$ For the OP's case, a=b=1, $f(x)=2qe^{-x} implies u(k)=2q, f(0)=2q, f(infty)=0.$ We will eventually show $$ (1) quad g(x)=frac{sinh(q x)}{sinh(x/2)} = -sum_{n=0}^infty frac{(-x)^n}{n!} cos(pi n) big( zeta(-n, 1/2+q) - zeta(-n, 1/2-q) big)$$ where $zeta(s,a)$ is the Hurwitz zeta function. Given (1), it is easy to see that $$ frac{d}{ds} log{v(s)} big|_{s=0} = frac{v'(0)}{v(0)} = -frac{ zeta'(0, 1/2+q)-zeta'(0, 1/2-q)}{ zeta(0, 1/2+q)-zeta(0, 1/2-q) }$$ However, it is known that $$zeta'(0,a)=log(Gamma(a)/sqrt{2pi}) text{ and } zeta(0,a)=-B_1(a)=1/2-a $$ where in the last formula the Hurwitz zeta has been connected to the Bernoulli polynomial by the formula $$ (2) quad zeta(-n,a) = -frac{B_{n+1}(a)}{n+1}. $$ Doing the rest of the algebra results in the expression $$ (3) quad int_0^infty Big(2qe^{-x} - frac{ sinh(q x)}{sinh{x/2} } Big) frac{dx}{x} = log{Gamma(1/2+q)} - log{Gamma(1/2-q)}. $$ To get it in the form of the OP's request, use the gamma-function reflection formula
$$ Gamma(1/2-q)Gamma(1/2+q) = frac{pi}{cos{pi q } } .$$
Now, to prove (1): $$ frac{sinh(q x)}{sinh(x/2)} = frac{e^{qx} - q^{-qx}}{e^{x/2}-e^{-x/2}} = frac{1}{x}Big( frac{x}{e^x-1} exp(x(1/2+q))+frac{x}{e^x-1} exp(x(1/2-q)) Big) $$ Use the well-known generating function for Bernoulli polynomials, $$ frac{sinh(q x)}{sinh(x/2)} =frac{1}{x}sum_{n=0}^infty frac{x^n}{n!} Big( B_n(1/2+q) - B_n(1/2-q) Big) =sum_{n=0}^infty frac{x^n}{n!(n+1)} Big( B_{n+1}(1/2+q) - B_{n+1}(1/2-q) Big) $$ where in the second step we have re-indexed because $B_0(x)=1$ and the first term is thus zero. Then use (2) in the last formula to complete the proof of (1).
Correct answer by skbmoore on December 3, 2020
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
With $ds{q in mathbb{R}}$, note that
begin{align}
&bbox[10px,#ffd]{left.int_{0}^{infty}{1 over x}bracks{2qexpo{-x} -
{sinhpars{qx} over sinhpars{x/2}}}dd x
,rightvert_{{large q in mathbb{R}} atop
{large q^{2} < 1/4}}}
\[5mm] = &
left.mrm{sgn}pars{q}int_{0}^{infty}{1 over x}
bracks{2verts{q}expo{-x} -
{sinhpars{verts{q}x} over sinhpars{x/2}}}dd x
,rightvert_{ verts{q} < 1/2}
label{1}tag{1}
end{align}
The last integral becomes:
begin{align}
&!!!!!!!!!!bbox[10px,#ffd]{int_{0}^{infty}{1 over x}bracks{2verts{q}expo{-x} -
{expo{-pars{1/2 - verts{q}}x} - expo{-pars{1/2 + verts{q}}x} over 1 - expo{-x}}}dd x}
\[5mm] stackrel{x = -lnpars{t}}{=} &
int_{1}^{0}{1 over -lnpars{t}}pars{2verts{q}t -
{t^{1/2 - verts{q}} - t^{1/2 + verts{q}} over 1 - t}}
pars{-,{dd t over t}}
\[5mm] = & !!!!
int_{0}^{1}!!underbrace{bracks{-,{1 over lnpars{t}}}}
_{ds{int_{1}^{infty}t^{xi - 1},ddxi}}
{2verts{q} - 2verts{q}t - t^{-1/2 - verts{q}} + t^{-1/2 + verts{q}}
over 1 - t},
dd t
\[5mm] = &!!!
int_{1}^{infty}!!!!int_{0}^{1}!!
{2verts{q}t^{xi - 1} - 2verts{q}t^{xi} - t^{xi - 3/2 - verts{q}} + t^{xi - 3/2 + verts{q}} over 1 - t},dd t,ddxi
\[5mm] = &
int_{1}^{infty}int_{0}^{1}left[%
2verts{q}int_{0}^{1}{1 - t^{xi} over 1 - t},dd t -
2verts{q}int_{0}^{1}{1 - t^{xi - 1} over 1 - t},dd tright.
\[2mm] & left. +int_{0}^{1}{1 - t^{xi - 3/2 - verts{q}} over 1 - t},dd t -
int_{0}^{1}{1 - t^{xi - 3/2 + verts{q}} over 1 - t},dd t
right]ddxi
\[5mm] = &
int_{1}^{infty}left[vphantom{huge A},%
2verts{q}Psipars{xi + 1} - 2verts{q}Psipars{xi}right.
\[2mm] & phantom{int_{1}^{infty}left[right.}
left. + Psipars{xi - {1 over 2} - verts{q}} -
Psipars{xi - {1 over 2} + verts{q}}right]ddxi
label{2}tag{2}
\[5mm] = &
overbrace{lim_{xi to infty}bracks{
2verts{q}lnpars{xi} + lnpars{Gammapars{xi - 1/2 - verts{q}} over Gammapars{xi - 1/2 + verts{q}}}}}
^{ds{= 0}}
\[2mm] &
+lnpars{Gammapars{1/2 + verts{q}} over Gammapars{1/2 - verts{q}}}
\[5mm] = &
lnpars{Gamma^{, 2}pars{1/2 + verts{q}} over
Gammapars{1/2 + verts{q}}Gammapars{1/2 - verts{q}}}
\[5mm] = &
lnpars{Gamma^{, 2}pars{1/2 + verts{q}} over
pi/sinpars{pibracks{1/2 + q}}}label{3}tag{3}
\[5mm] = &
bbx{lnpars{cospars{piverts{q}}} +
2lnpars{Gammapars{verts{q} + {1 over 2}}}
- lnpars{pi}}label{4}tag{4} \ &
end{align}
(ref{2}): I used an integral representation
( see $ds{color{black}{bf 6.3.22}}$ in A & S Table ) of the digamma $ds{mbox{function} Psi}$.
(ref{3}): Euler Reflection Formula. See
$ds{color{black}{bf 6.1.17}}$ in A & S Table.
Answered by Felix Marin on December 3, 2020
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