Prove Sum Equals Catalan's Constant

Question

How do I prove the following identity

$$G = \frac{\pi \log(2)}{4} - \frac{\sqrt{\pi}}{4}\sum_{k=1}^{\infty} \frac{(-1)^k \Gamma\left(\frac{1+k}{2}\right)}{k \Gamma\left(1 + \frac{k}{2}\right)}$$

where $G$ is Catalan's Constant, and $\Gamma(x)$ is the Gamma Function. This problem was given to me by a friend as a challenge, but I am unable to prove the relation above. I verified this using Mathematica, this computational difference between two sides of the equality is $(2.77556)10^{-16}$, so I suspect the identity is correct. I attempted using the following properties of the Gamma function

$$\Gamma(1 + z) = z \Gamma(z)$$ $$\Gamma\left(\frac{1}{2} + n\right) = \frac{(2n)! \sqrt{\pi}}{4^n n}$$

Yet this did not help my evaluation resulting summation. Any input or answer would be greatly appreciated.

Answer 1

Split up the summand

$$a_n = \frac{(-1)^n \Gamma\left(\frac12+\frac n2\right)}{n\,\Gamma\left(1+\frac n2\right)}$$

into cases of even and odd indices:

$$\begin{cases}\displaystyle a_{2k} = \frac{\Gamma\left(\frac12+k\right)}{2k \, \Gamma(1+k)} = \frac{\sqrt\pi}{2^{2k+1} k} \binom{2k}k \\ \displaystyle a_{2k-1} = -\frac{\Gamma(k)}{(2k-1)\,\Gamma\left(\frac12+k\right)} = -\frac{2^{2k}}{\sqrt\pi \, k(2k-1) \binom{2k}k}\end{cases}$$

For the even case, recall and manipulate the binomial series:

$$\begin{align*} \frac1{\sqrt{1-x^2}} &= \sum_{k\ge0} \frac1{2^{2k}} \binom{2k}k x^{2k} \\ \implies \frac1{x\sqrt{1-x^2}} &= \frac1x + \sum_{k\ge1} \frac1{2^{2k}} \binom{2k}k x^{2k-1} \\ \implies \sum_{k\ge1} \frac1{2^{2k} k} \binom{2k}k x^{2k} &= C - \log\left(1+\sqrt{1-x^2}\right) \end{align*}$$

where $C$ is a constant picked up from integrating. Notice that $x=0\implies C=\log2$, so

$$\sum_{k\ge1} a_{2k} x^{2k} = \sqrt\pi \log\frac2{1+\sqrt{1-x^2}} \implies \boxed{\sum_{k\ge1} a_{2k} = \sqrt\pi \log2}$$

For the odd case, use the series expansion of $\arcsin^2x$:

$$\begin{align*} \arcsin^2x &= \sum_{k\ge1} \frac{2^{2k-1}}{k^2 \binom{2k}k} x^{2k} \\ \implies \frac{\arcsin x}{x\sqrt{1-x^2}} &= \sum_{k\ge1} \frac{2^{2k-1}}{k \binom{2k}k} x^{2k-2} \\ \implies C + \int_0^x \frac{\arcsin t}{t\sqrt{1-t^2}} \, dt &= \sum_{k\ge1} \frac{2^{2k-1}}{k (2k-1) \binom{2k}k} x^{2k-1} \end{align*}$$

Now $x=0\implies C=0$, and as $x\to1^-$ the integral on the LHS converges to a known representation of Catalan's constant,

$$\int_0^1 \frac{\arcsin t}{t\sqrt{1-t^2}} \, dt \stackrel{t=\tfrac u{\sqrt{1+u^2}}}= \int_0^\infty \frac{\arctan u}{u\sqrt{1+u^2}} \, du = 2G$$

and it follows that

$$\sum_{k\ge1} a_{2k-1} x^{2k-1} = -\frac2{\sqrt\pi} \int_0^x \frac{\arcsin t}{t\sqrt{1-t^2}} \, dt \implies \boxed{\sum_{k\ge1} a_{2k-1} = -\frac{4G}{\sqrt\pi}}$$

Answer 2

We use the beta function. Let $$ S = \sum_{k=1}^{\infty} \frac{(-1)^k}{k}\frac{\Gamma\! \left(\tfrac{k+1}{2} \right)} {\Gamma\! \left(1 + \tfrac{k}{2} \right)}. $$

We have $$\frac{\Gamma(\frac{1+k}{2})\Gamma(\frac{1}{2})}{\Gamma(1 + \frac{k}{2})\Gamma(\frac{1}{2})} = \frac{1}{\sqrt{\pi}}B(\frac{k+1}{2},\frac{1}{2}),$$

so by definition of the beta function, we can rewrite $S$ as

$$ S= \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \cdot \frac{1}{\sqrt{\pi}} \int_{0}^{1} t^{\tfrac{k-1}{2}}(1-t)^{-\tfrac12}dt. $$

Interchanging sum and integral: $$ S = \frac{1}{\sqrt{\pi}} \int_{0}^{1}(1-t)^{-\tfrac12}\left[ \sum_{k=1}^\infty \frac{(-1)^k}{k}t^{\tfrac{k-1}{2}} \right] dt. $$ Now, using $\sum_{k=0}^\infty \frac{x^k}{k} = -\ln(1-x)$, with $x=-\sqrt t$, we have $$ \sum_{k=1}^\infty \frac{(-1)^k}{k}t^{\tfrac{k-1}{2}} = t^{-\tfrac12}\sum_{k=1}^\infty \frac{(-\sqrt{t})^k}{k} = t^{-\tfrac12} \left[-\ln(1 + \sqrt{t}) \right] = -\frac{\ln(1+\sqrt{t})}{\sqrt{t}}. $$ Hence $$ S = \frac{1}{\sqrt{\pi}} \int_{0}^{1} (1-t)^{-\tfrac12} \left[-t^{-\tfrac12}\ln \left(1+\sqrt{t} \right) \right]dt = - \frac{1}{\sqrt{\pi}} \int_{0}^{1} \frac{\ln \left(1+\sqrt{t} \right)}{\sqrt{t(1-t)}}dt. $$

Let us now transform this integral into a more recognizable form. Use the substitution $t = \sin^2\theta$ for $\theta \in [0,\tfrac{\pi}{2}]$, then $$ dt = 2 \sin\theta\cos\theta d\theta, \quad \sqrt{t} = \sin\theta, \quad \sqrt{1-t} = \cos\theta. $$ Hence $$ \int_0^1 \frac{\ln(1+\sqrt{t})}{\sqrt{t(1-t)}} dt = \int_{0}^{\pi/2} \frac{\ln \left(1+\sin\theta \right)} {\sin\theta\cos\theta} 2\sin\theta\cos\theta d\theta = 2\int_{0}^{\pi/2}\ln \left(1+\sin\theta \right) d\theta. $$ Now, $\ln(\sec \theta+\tan\theta) = \ln(1+\sin\theta)-\ln(\cos\theta)$, so

$$ \int_{0}^{\pi/2}\ln \left(1+\sin\theta \right) d\theta = \int_{0}^{\pi/2}\ln(\sec \theta+\tan\theta) d\theta+\int_{0}^{\pi/2}\ln(\cos\theta) d\theta. $$

The former summand is (according to Wikipedia) $2G$, and the latter is the famous "logsine" integral of value $-\frac{\pi}{2}\ln 2$. Putting this all together, we have $$ S = - \frac{2}{\sqrt{\pi}} \int_{0}^{\pi/2}\ln \left(1+\sin\theta \right) d\theta = - \frac{2}{\sqrt{\pi}}\left(2G-\frac{\pi}{2}\ln 2\right). $$ Rearranging, we obtain $$ G = \frac{\pi\ln 2}{4}-\frac{\sqrt\pi}{4}S, $$ as desired.

This argument can likely be made easier using some trigonometric definition of the beta function.

Answer 3

\begin{align*}\frac{\pi \log(2)}{4} - \frac{\sqrt{\pi}}{4}\sum_{k=1}^{\infty} \frac{(-1)^k \Gamma\left(\frac{1+k}{2}\right)}{k \Gamma\left(1 + \frac{k}{2}\right)} &= \frac{\pi \log(2)}{4} +\frac{1}{2}\sum_{k=1}^{\infty} \frac{(-1)^{k+1}\text{B}\left({\frac{n+1}{2},\frac{1}{2}}\right)}{2k } \\ &= \frac{\pi \log(2)}{4}+\frac{1}{2}\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k }\int^{\pi/2}_0\sin^k(x)dx\\&= \frac{\pi \log(2)}{4}+\frac{1}{2}\int^{\pi/2}_0 \ln(1+\sin(x))dx\\&=\text{G}\end{align*}