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

Question

Gradshteyn&Ryzhik $3.554.5$ states that:
$$\int_0^{\infty } \frac1x \biggl( 2qe^{-x}-\frac{\sinh (q x)}{\sinh \left(\frac{x}{2}\right)} \biggr) \, dx=\log\bigl(\cos (\pi q) \bigr)+2 \log \left(\Gamma \bigl(q+\frac12 \bigr)\right)-\log (\pi )~, \quad q^2<\frac{1}{4}$$
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.

Answer 1

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 $$\zeta'(0,a)=\log(\Gamma(a)/\sqrt{2\pi}) \text{ and } \zeta(0,a)=-B_1(a)=1/2-a $$ 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).