How to prove this gamma function identity?

Reading Landau and Lifshitz "Quantum Mechanics. Non-relativistic theory", I've come across an identity, which after being a bit simplified, reads

$$\left|\frac{((a+b)\Gamma(2a)\Gamma(-(a+b))^2}{(a-b)\Gamma(-2a)\Gamma(a-b)^2}\right|=\left|\frac{\sin(\pi(a-b))}{\sin(\pi(a+b))}\right|,\tag1$$

and apparently holds for pure imaginary $a$ and $b$. The book says the left hand side can be calculated to equal right hand side using the reflection formula:

$$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}.\tag2$$

But I fail to see how I could actually do this calculation. Plotting the difference I see that this actually is false for real $a$ and $b$, so the reflection formula, which AFAIK works for all $x\in\mathbb C$ seems to be not enough to get this result.

So, how do I prove $(1)$, or even better, derive RHS from LHS?

The identity holds for pure imaginary $a$ and $b$. So, let $a=i u$ and $b=i v$ where $u$ and $v$ are real. The identity to be proved is : $$\left|\frac{(i(u+v)\Gamma(2iu)\Gamma(-i(u+v))^2}{i(u-v)\Gamma(-2iu)\Gamma(i(u-v))^2}\right|=\left|\frac{\sin(\pi i(u-v))}{\sin(\pi i(u+v))}\right|$$
At first look, I don't see any interest to use the reflexion relationship. It is much easier to use the formula giving the modulus : $$|\Gamma{(iy)}|=\sqrt{\frac{\pi}{y \sinh(\pi y)}}$$ With this formula, remplace $y$ by $(u+v)$ or $(u-v)$ and put it back into the left term above. The simplification leads to $\frac{\sinh(\pi(u-v))}{\sinh(\pi(u+v))}$
With $\sinh(y)=-i\sin(i y)$ we have: $\frac{\sinh(\pi(u-v))}{\sinh(\pi(u+v))}=\frac{\sin(\pi i(u-v))}{\sin(\pi i(u+v))}$ which is the right term of the equation and so, it is proved.
• Wow! In fact the original variables were indeed sum and difference multiplied by $i$, it was me who changed them into $a$ and $b$ trying to work out the result. – Ruslan Nov 18 '14 at 19:19
• You seem to have lost a minus in $\sinh(y)=-i\sin(iy)$. – Ruslan Nov 18 '14 at 19:47