Proving $\sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a)$

Let's transform it into a factorial form

$\sum_\limits{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=\sum_\limits{n=0}^\infty\frac{(-1)^n(2n+a)!a!}{(2n+1)!a!}=\sum_\limits{n=0}^\infty\frac{(i)^{2n}\binom{2n+a}{2n}a!}{2n+1}$

After reindexing and using the mentioned series property we have:

$\sum_\limits{n=0}^\infty\frac{(i)^{2n}\binom{2n+a}{2n}a!}{2n+1}=\Re \sum_\limits{n=0}^\infty\frac{(i)^{n}\binom{n+a}{n}a!}{n+1}$

Apply the fact $\frac{1}{n+1}=\int\limits_0^1 t^n dt$ and the binomial identity: $\frac{1}{(1-z)^{a+1}}=\sum\limits_{n=0}^\infty \binom{n+a}{n}z^a$

We get:

$\Re \int\limits_0^1\frac{\Gamma(a+1)}{(1- it)^{a+1}}dt=\Re \int\limits_{1-i}^1\frac{\Gamma(a+1)}{ix^{a+1}}dx=\frac {\Gamma(a+1)}{a}\sqrt{2}^{-a}\sin(\frac{\pi}{4}a)$

A solution in large steps by Cornel Ioan Valean

In the following, I'll focus on the last series. Let's prove that

$$\sum_{n=1}^{\infty} x^n \frac{\Gamma(n+a-1)}{\Gamma(n)}=\frac{x}{(1-x)^a}\Gamma(a).$$

Two key steps are necessary:

$1)$. Note and use that

$$\frac{1}{\Gamma(1-a)}\int_0^1 t^{-a} (1-t)^{n+a-2}\textrm{d}t=\frac{\Gamma(n+a-1)}{\Gamma(n)}.$$

$2)$. (after summing) Employ the following integral representation with a hypergeometric structure (in fact, it may be viewed as a particular case of an integral expressed in terms of a hypergeometric function)

$$\int_0^1 \frac{x^{a-1}}{(1-x)^a (1+b x)}\textrm{d}x=\frac{\pi}{\sin(\pi a)}\frac{1}{(1+b)^a}.$$

One useful way to perform the evaluation of the last integral is by using the variable change $x/(1-x)=y$, followed by the variable change $(1+b)y=z$ in order to get precisely a special case of the Beta function.

End of story