Calculation of a derivative of a function related to the Euler Gamma function

Question

Let $F$ be defined by $ F(x)=\frac{\Gamma(\frac{1 + x}2)}{\sqrt π\ \Gamma(1 + \frac x2)}. $ I believe that $F'(0)=-\ln 2$, but I do not have a proof. Is there an easy way to get that result?

Answer 1

Set $f(x) = \Gamma(\frac{1+x}{2})$, $g(x) = \sqrt{\pi}\Gamma(1+\frac{x}{2})$. We compute the derivatives of $f$ and $g$ to get $$f'(x) = \frac{1}{2}\int_0^\infty t^{\frac{x-1}{2}}e^{-t}\ln(t)\, dt \quad \mathrm{and} \quad g'(x) = \frac{\sqrt{\pi}}{2}\int_0^\infty t^{\frac{x}{2}}e^{-t}\ln(t)\, dt.$$ Now, by the quotient rule we have $$F'(0) = \frac{ \left[\frac{\sqrt{\pi} \, \Gamma(1)}{2}\int_0^\infty t^{-1/2}e^{-t}\ln(t)\, dt \right] - \left[ \frac{\sqrt{\pi}\, \Gamma(1/2)}{2}\int_0^\infty e^{-t}\ln(t)\, dt \right]}{\pi\, \Gamma(1)^2}.$$ Wolfram alpha evaluates the first integral as $-\sqrt{\pi}(\gamma + \ln(4))$, and the second integral is known to be $-\gamma$ (here, $\gamma$ is the Euler-Mascheroni constant). Hence $$ F'(0) = \frac{- \frac{\pi}{2}(\gamma + \ln(4)) + \frac{\pi}{2}\gamma}{\pi} = -\ln(4)/2 = -\ln(2).$$

Answer 2

Too long for a comment. I am grateful to Inbo Gottlieb-Fenves for the nice answer. I am now realizing that I should have said that $ F(x)=\frac2π\int_0^{π/2}(\cos \theta)^x d\theta, $ so that $$ \frac π2F'(0)=\int_0^{π/2}\ln(\cos \theta) d\theta=\int_0^{π/2}\ln(\sin \theta) d\theta=\int_0^{π/2}\ln(\sin 2\theta) d\theta, $$ and thus $ πF'(0)=\int_0^{π/2}\ln(\cos \theta \sin \theta) d\theta= \int_0^{π/2}\ln(\frac12\sin 2\theta) d\theta=-\fracπ2\ln 2+\fracπ2F'(0), $ implying $$ F'(0)=-\ln 2.\hskip55pt \square $$

Answer 3

The Taylor series of $F(x)$ could be a solution for even higher derivatives.

Let $x=2t$ and use $$\log \big(\Gamma (a+t)\big)=\sum_{n=0}^\infty \frac{\psi ^{(n-1)}(a)}{n!}\,t^n$$ $$\log \left(\Gamma \left(\frac{1}{2}+t\right)\right)-\log \Big(\Gamma \left(1+t\right)\Big)=\sum_{n=0}^\infty \frac{\psi ^{(n-1)}\left(\frac{1}{2}\right)-\psi ^{(n-1)}(1)}{n!}\,t^n$$

If $$b_n=\frac{\psi ^{(n-1)}\left(\frac{1}{2}\right)-\psi ^{(n-1)}(1)}{n!}$$ the first of them are $$\left\{\frac{\log (\pi )}{2},-2\log (2),\frac{\pi ^2}{6},-2 \zeta (3),\frac{7 \pi ^4}{180},-6 \zeta (5),\frac{31 \pi ^6}{2835}\right\}$$ Now, continue with Taylor series using $$\frac 1 {\sqrt \pi}\,\frac{\Gamma \left(\frac{1}{2}+t\right)}{\Gamma (1+t)}=\frac 1 {\sqrt \pi}\,\exp\Bigg( \log \left(\Gamma \left(\frac{1}{2}+t\right)\right)-\log \Big(\Gamma \left(1+t\right)\Big)\Bigg)$$ to obtain $$1-2\log (2)\,t+\frac{1}{6} \left(\pi ^2+12 \log ^2(2)\right)\,t^2+\,\cdots$$

Replace $t$ by $\frac x2$ or use the chain rule.