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Evaluate $int_0^{infty } Bigl( 2qe^{-x}-frac{sinh (q x)}{sinh left(frac{x}{2}right)} Bigr) frac{dx}x$

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.

2 Answers

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.


Finally, with (ref{1}) and (ref{4}): 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] = & bbox[25px,#ffd,border:1px groove navy]{mrm{sgn}pars{q}bracks{lnpars{cospars{piverts{q}}} + 2lnpars{Gammapars{verts{q} + {1 over 2}}} - lnpars{pi}}} \ & end{align}

Answered by Felix Marin on December 3, 2020

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