Is there a simpler form for $\Re \frac{\Gamma(1/2-i)}{\Gamma(1-i)}$?

Is there a simpler (i.e. manifestly real) form for $\Re \frac{\Gamma(1/2-i)}{\Gamma(1-i)}$ or $\Im \frac{\Gamma(1/2-i)}{\Gamma(1-i)}$, or more generally for $\frac{\Gamma(1/2-ia)}{\Gamma(1-ia)}$ with $a \in \mathbb{R}$? This question arose in the evaluation of the integral $$\int_0^{\pi/2}\cos\left(2a \ln \sin x \right) \, dx= \frac{\sqrt \pi}{2}\Re \frac{\Gamma(1/2-ia)}{\Gamma(1-ia)}$$

After 'FullSimplify'-ing, Mathematica gives (for $a=1$)

$$\frac{1}{\pi }2^{-3-2 i} \text{Sinh}[2 \pi ] \left(2^{2 i} \sqrt{\pi } \text{Csch}[\pi ] \left(\text{Gamma}\left[\frac{1}{2}+i\right] \text{Gamma}[1-i]-\text{Gamma}\left[\frac{1}{2}-i\right] \text{Gamma}[1+i]\right)+\text{Gamma}[1-i]^2 \text{Gamma}[1+2 i] (-1+\text{Tanh}[\pi ])+2^{4 i} \text{Gamma}[1+i]^2 \text{Gamma}[1-2 i] (1+\text{Tanh}[\pi ])\right)$$ which I don't find simpler at all and still contain's $i$'s.

I tried to use the product representation for $\Gamma$ and $1/\Gamma$ but those attempts did not result in anything useful.

• $i$ is the imaginary unit. – user111187 May 10 '14 at 14:18

No, one only has a simpler formula for the absolute value, since $$\left|\frac{\Gamma(1/2-i)}{\Gamma(1-i)}\right|^2=\frac{\Gamma(1/2-i)}{\Gamma(1-i)}\frac{\Gamma(1/2+i)}{\Gamma(1+i)}=\frac{\sin\pi i }{i\sin\pi(1/2-i)}=\tanh\pi.$$
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