How would you solve for $x$ in the following equation:
$$ x = \left( x - 1 \right) ! = \int_0^\infty t^{x-1} e^{-t} dt $$
If we are only concerned about integers, then clearly, the only solution is $1$. But if we graph each side, we can see that there is one more positive solution (that is not an integer) and an infinite amount of negative solutions, two between each interval of length one where the endpoint of each is a negative even integer, beginning with $\left( -3, -2 \right)$ and going off to negative infinity. I'm actually more concerned with the non-integer positive solution. I'm just curious without graphing it, how would one go about finding the other solutions, mainly the other positive solution (which is $x \approx 3.562$)?
To generalize this even further, how would you go about finding the other positive solution for a constant $k > 0$ in:
$$ kx = \left( x - 1 \right) ! = \int_0^\infty t^{x-1} e^{-t} dt $$
Although, I do realize if $k$ is less than approximately $0.4983$, then there is no positive solution. I'm very interested in seeing how this is done. I'd appreciate any feedback.