Is there any way to simplify the following expression, $$\frac{\Gamma\left(x\right)^{2}}{\Gamma\left(x-\sqrt{2}\right)\Gamma\left(x+\sqrt{2}\right)}$$ This actually prompts me to ask whether we have any identity for $$\Gamma\left(a-b\right)\Gamma\left(a+b\right)?$$ I know there's an identity for $\Gamma\left(a-b\right)\Gamma\left(b-a\right)$ via the reflection-formula. The first equation can be expressed as in terms of the Beta function, but I don't want to unnecessarily bring in that. So will this stay as it is or can it be a little bit more simplified? Also $x\in\mathbb R$
Some elementary simplifications I could find is, $$-2\operatorname{sinc}\left(\pi\sqrt{2}\right)\cdot \operatorname{B}\left(x,\sqrt{2}\right)\operatorname{B}\left(x,-\sqrt{2}\right)$$
Well I guess that's it.
You do not have much choice for rewriting this expression.
However, you can have excellent approximation using series or Padé approximants even for small values of $x$. For example $$\frac{\Gamma (x)^2}{\Gamma (x-a) \Gamma (x+a)}=1-\frac{a^2}{x}+\frac{a^2 \left(a^2-1\right)}{2 x^2}-\frac{a^2 \left(a^2-1\right)^2}{6 x^3}+\frac{a^4 \left(a^2-1\right)^2}{24 x^4}-\frac{a^2 (a^2-1)^2 (a^4+2a^2-4) }{120 x^5}+O\left(\frac{1}{x^6}\right)$$
For $a=\sqrt 2$, this gives an absolute error of $4.87 \times 10^{-6}$ for $x=4$ (relative error : $8.72\times 10^{-4}$%)
Edit
Writing $$\frac{\Gamma (x)^2}{\Gamma (x-a) \Gamma (x+a)}=1-\frac{a^2}{x}+\sum_{n=2}^\infty \frac {c_n}{x^n}$$ let $$d_n=(-1)^n {n!}\,c_n$$ the $d_n$'s are polynomials in $a^2$.
There is a different pattern depending on the parity.
Let $b=a^2$ and here are the first terms $$\left( \begin{array}{cc} n & P_n(b) \\ 3 & 1 \\ 5 & b^2+2 b-4 \\ 7 & b^4+16 b^3-11 b^2-96 b+120 \\ 9 & b^6+50 b^5+393 b^4-1660 b^3-1520 b^2+11808 b-12096 \end{array} \right)$$
$$\left( \begin{array}{cc} n & Q_n(b) \\ 4 & 1 \\ 6 & b^2+7 b-14 \\ 8 & b^4+30 b^3+73 b^2-600 b+736 \\ 10 & b^6+77 b^5+1239 b^4-1129 b^3-30716 b^2+104832 b-104544 \end{array} \right)$$
Using all the above for $a=\sqrt 2$ and $x=4$, the absolute error is $2.40\times 10^{-10}$ (relative error : $4.31\times 10^{-8}$%).
Now, what are these polynomials ? "That is the question !".
Edit
It seems that it could be better to consider the reciprocal of the expression $$\frac{\Gamma (x-a) \Gamma (x+a)}{\Gamma (x)^2}=1+\frac{a^2}{x}+\frac{a^2 \left(a^2+1\right)}{2 x^2}+\frac{a^2 \left(a^4+4 a^2+1\right)}{6 x^3}+\frac{a^4 \left(a^4+10 a^2+13\right)}{24 x^4}+\frac{a^2 \left(a^8+20 a^6+73 a^4+30 a^2-4\right)}{120 x^5}+O\left(\frac{1}{x^6}\right)$$ Writing $$\frac{\Gamma (x-a) \Gamma (x+a)}{\Gamma (x)^2}=+\frac{a^2}{x}+\frac{a^2 \left(a^2+1\right)}{2 x^2}+\sum_{n=3}^\infty \frac {c_n}{x^n}$$ let $$d_n={n!}\,c_n$$ the $d_n$'s are polynomials in $a^2$.
There is a different pattern depending on the parity.
Let $b=a^2$ and here are the first terms $$\left( \begin{array}{cc} n & P_n(b) \\ 3 & b^2+4 b+1 \\ 5 & b^4+20 b^3+73 b^2+30 b-4 \\ 7 & b^6+56 b^5+798 b^4+3008 b^3+1561 b^2-504 b+120 \end{array} \right)$$
$$\left( \begin{array}{cc} n & Q_n(b) \\ 4 & b^2+10 b+13 \\ 6 & b^4+35 b^3+273 b^2+425 b-14 \\ 8 & b^6+84 b^5+1974 b^4+14572 b^3+25809 b^2-2856 b+736 \end{array} \right)$$