Sum on non-integer gamma argument

We note that $\begin{aligned} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \frac{\Gamma\big(\tfrac12n\big)}{\Gamma\big(\tfrac12(n+1)\big)} & = \; \frac{1}{\sqrt{\pi}}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} B\big(\tfrac12n,\tfrac12\big) \; = \; \frac{1}{\sqrt{\pi}}\sum_{n=1}^\infty \int_0^1 \frac{(-1)^{n+1}}{n}t^{\frac12n - 1}(1-t)^{-\frac12}\,dt \\ & = \; \frac{1}{\sqrt{\pi}}\int_0^1 \frac{\ln(1 + \sqrt{t})}{t\sqrt{1-t}}\,dt \; = \; \frac{2}{\sqrt{\pi}} \int_0^1 \frac{\ln(1+u)}{u\sqrt{1-u^2}}\,du \; =\; \frac{2}{\sqrt{\pi}}\int_0^{\frac12\pi} \frac{\ln(1 + \sin\theta)}{\sin\theta}\,d\theta \\ & = \; \frac{2}{\sqrt{\pi}} \times \tfrac18\pi^2 \; = \; \frac{\pi^{\frac32}}{4} \end{aligned}$ using the substitutions $t = u^2$ , $u = \sin\theta$ , and the calculations here to evaluate the final integral $\begin{aligned} \int_0^{\frac12\pi}\frac{\ln(1 + \sin\theta)}{\sin\theta}\,d\theta & =\; \int_0^1 \left(\int_0^{\frac12\pi}\frac{d\theta}{1 + a\sin\theta}\right)\,da \; = \; \int_0^1 \frac{2}{\sqrt{1-a^2}}\tan^{-1}\sqrt{\frac{1-a}{1+a}} \,da \\ & = \; 2\int_{\frac12\pi}^0 \frac{1}{\sin\theta}\tan^{-1}\sqrt{\frac{2\sin^2\frac12\theta}{2\cos^2\frac12\theta}}\,(-\sin\theta)\,d\theta \; = \; \tfrac18\pi^2 \end{aligned}$ making the answer $3 + 2 + 4 = \boxed{9}$ .

First we establish that $\int_0^{1}\frac{\ln(x+\sqrt{1-x^2})}{x}dx=\frac{\sqrt{\pi}}{4}{\color{#20A900}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}}$ So we proceed by substituting $x=\cos y$ , then we have $\int_0^{\frac{\pi}{2}}\frac{\ln\left(x+\sqrt{1-x^2}\right)}{x}dx=\int_0^{\frac{\pi}{2}}\frac{\ln\left(\cos y+\sin y\right)}{\cos y}\sin ydy$ make the change of variable $y\mapsto \frac{\pi}{2} -y$ and hence of adding the obtained integral with latter integral, we get $\int_0^{\frac{\pi}{2}}\frac{\ln\left(\cos y+\sin y\right)}{\cos y}\sin ydy=\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\ln\left(1+\sin y\right)}{\sin y}dy$ For $0 < y< \frac{\pi}{2}$ , $0< \sin y <1$ and hence last integral reduces to( by series expansion ) $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n}\int_0^{\frac{\pi}{2}}\sin ^{n-1} y dy\overset{\text{Wallis' Int}}{=}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{4n}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{1}{2}+\frac{n}{2}\right)}\\ \hspace{4.75cm}=\frac{\sqrt{\pi}}{4}{\color{#20A900}{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}}}$ Now split the sum into even and odd parity, giving us $\sum_{n=1}^{\infty}\frac{\Gamma\left(n-\frac{1}{2}\right)}{(2n-1)\Gamma(n)}-\sum_{n=1}^{\infty}\frac{\Gamma(n)}{2n\Gamma\left(n+\frac{1}{2}\right)}$ Now, we evaluate the former sum $\sum_{n=1}^{\infty}\frac{\Gamma\left(n-\frac{1}{2}\right)}{(2n-1)\Gamma(n)}=\sum_{n=0}^{\infty}\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)\Gamma(n+1)}=\sqrt{\pi}{\color{#D61F06}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)4^n}{2n\choose n}}}=\pi^{1/2}{\color{#D61F06}{\left[\frac{\pi}{2}\right]}}$ By generating function of central binomial coefficients we $\displaystyle \sum_{n\geq 0}\frac{x^n}{4^n}{2n\choose n}=\frac{1}{\sqrt{1-x}}$ . Replacing $x$ by $x^2$ on integrating from $0$ to $1$ gives ${\color{#D61F06}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)4^n}{2n\choose n}}} =\int_0^1\frac{dx}{\sqrt{1-x^2}}={\color{#D61F06}{\frac{\pi}{2}}}$ Further, we evaluate the latter sum $\sum_{n=1}^{\infty}\frac{\Gamma(n)}{2n\Gamma\left(n+\frac{1}{2}\right)}=\frac{\pi^{-1/2}}{2}{\color{#3D99F6}\sum_{n=1}^{\infty}\frac{4^n}{n^2}{2n\choose n}^{-1}}=\frac{\pi^{-1/2}}{2}{\color{#3D99F6}\left[\frac{\pi^2}{2}\right]}$ we use the generating function $\displaystyle \arcsin^2(x)=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$ ( see here ). Set $x=1$ we have our answer $\frac{\pi^2}{4}$ . Therefore our final answer $\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}=\frac{\pi^{3/2}}{2}-\frac{\pi^{3/2}}{4}=\frac{\pi^{3/2}}{4}$ On multiply by $\frac{\pi^{1/2}}{4}$ gives us

$\int_0^1\frac{\ln\left(x+\sqrt{1-x^2}\right)}{x}dx=\frac{\pi^2}{16}$

If one is curious to derive the cited generating function then it quite easy to see by Lehmer identity, AMM, 1985 $\sum_{m\geq 1}\frac{(2x)^{2m}}{m{2m\choose m}}=\frac{2x\arcsin(x)}{\sqrt{1-x^2}}$ Dividing by by $x$ and hence on integrating gives us the desired form. For more interesting series on central binomial coefficients due to Lehmer, see here .