I can compute the gamma function when it is an integer multiple of $\frac{1}{2}$ or when it is a whole number. However, in this case $\frac{3}{4}$ is neither. How do I go about computing this?
I know the following:
$$\Gamma(n)=(n-1)!$$
$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$
$$\Gamma(n+1)=n\Gamma(n)$$
Without more context probably the best one can do is re-express it in terms of $\Gamma\left(\frac{1}{4}\right)$ using Euler's reflection formula:
$$\Gamma\left(\frac{3}{4}\right) \Gamma\left(1 - \frac{3}{4}\right) = \pi \csc \pi \left(\frac34\right),$$ which yields $$\Gamma\left(\frac{3}{4}\right) = \frac{\pi\sqrt{2}}{\Gamma\left(\frac14\right)} = 1.22541\ldots .$$
There's no known expression for $\Gamma\left(\frac14\right)$ in terms of more elementary functions and more familiar constants, so per the above equation the same must be true for $\Gamma\left(\frac34\right)$. As far as I know the same is true for $\Gamma(q)$ for every rational $q$ that is not a half-integer.
Incidentally, evaluating the approximation $$\Gamma(z) \approx \sqrt{2 \pi} z^{z - \frac{1}{2}} e^{-z} \left(1 + \frac{1}{12 z}\right),$$ which is essentially Stirling's approximation, at $z = \frac{3}{4}$ gives $$\Gamma\left(\frac34\right) \approx \frac{2^{1/2}3^{1/4} 4^{3/4} 5 \pi^{1/2}}{18e^{3 / 4}} = 1.22431\ldots,$$ which carries a relative error of less than $10^{-3}$.