Let define the function $x>0$:
$$f(x)=\Gamma\Big(\frac{1}{x}\Big)-x$$
Then $x_0$ such that :
$$f(x_0)=-\gamma$$
Define a local minima (see here) of the function $f(x)$.(We have the Gamma function and the negative Euler-Mascheroni constant)
To prove it I have tried derivative we have $x>0$ :
$$f'(x)=-\frac{\Gamma\Big(\frac{1}{x}\Big)\psi\Big(\frac{1}{x}\Big)}{x^2}-1$$
But we want :
$$f'(x)=0$$
Or :
$$\Gamma\Big(\frac{1}{x}\Big)\psi\Big(\frac{1}{x}\Big)=x^2$$
And now the best things I can do is to use Newton's method to check my result .
Question
Is there an hidden closed form here ?
How to solve it ?
Thanks in advance .