# Is there a function $f$ such that $\Gamma (c+x)=\Gamma (c-f\left( x \right) )$?

I was just looking at Euler's reflection formula for Gamma function which states $$\Gamma (1-z)\Gamma (z)=\frac { \pi }{ \sin { (z\pi ) } }$$ but it seems to me that one more reflection formula could exist so that $$\Gamma (c+x)=\Gamma (c-f\left( x \right) )?$$ where $c\approx 1.46163$ the local minimum for Gamma function where x>0.

Our first impressions would be that $$f\left( x \right) >f(y),\quad x>y\\ f(0)=0\\ f(2-c)=c-1\\ Im(f(x))=[0,c>$$.

Can such function exist, and still follow Euler's reflection formula?

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f(x) = -x ?? would work... – Gautam Shenoy Apr 12 '13 at 11:07
no, not even closely, the function must be defined on the whole $[0,+infinity>$ while it's image must be $Im(f(x))=[0,c>$ – user1398593 Apr 12 '13 at 11:11
it would be similar to arcus tangens function – user1398593 Apr 12 '13 at 17:46