This should've been mechanics!

If $\theta$ is the angle that the string of the pendulum makes with the downward vertical, then conservation of energy tells us that $\tfrac12 mL^2\dot\theta^2 + mgL(1 - \cos\theta) \;= \; mgL$ and hence that $\dot\theta^2 \; = \; \tfrac{2g}{L}\cos\theta$ Solving this differential equation, the period of the oscillation is $T \; = \; 2\int_{-\frac12\pi}^{\frac12\pi} \sqrt{\frac{L}{2g}} \frac{d\theta}{\sqrt{\cos\theta}} \; = \; \sqrt{\frac{8L}{g}}\int_0^{\frac12\pi} \frac{d\theta}{\sqrt{\cos\theta}} \; = \; \sqrt{\frac{8L}{g}}\int_0^{\frac12\pi} \frac{\sin\theta\,d\theta}{\sqrt{\cos\theta(1 - \cos^2\theta)}}$ With the substitution $u = \cos^2\theta$ this becomes $T \; = \; \sqrt{\frac{2L}{g}}\int_0^1 u^{-\frac34}(1-u)^{-\frac12}\,du \; = \; \sqrt{\frac{2L}{g}}B(\tfrac14,\tfrac12)$ Since $\Gamma(\tfrac12) = \sqrt{\pi}$ and $\Gamma(\tfrac14)\Gamma(\tfrac34) \,=\, \frac{\pi}{\sin\frac14\pi} \,=\, \pi\sqrt{2}$ , it follows that $T \; = \; \sqrt{\frac{L}{\pi g}}\Gamma(\tfrac14)^2$ making the answer $1 + 1 + 1 + 4^2 = \boxed{19}$ .

This all comes down to conservation of energy. We have:

$\frac{1}{2} mv^2 \ = \ mgL \cos \theta \ \Rightarrow \ v(\theta) \ = \ \sqrt{2gL \cos\ \theta} \ \Rightarrow \ \frac{1}{v(\theta)} \ = \ \frac{1}{\sqrt{2gL \cos\ \theta}}$

In order to get the period, we have to evaluate the following integral:

$T \ = \displaystyle\int_{s_1}^{s_2} \ \frac{1}{v(s)} \ ds$

We do a change of variables to $\theta$ . Since $\theta \ = \ \frac{s}{L} \ \Rightarrow \ s \ = \ \theta L$ , we get:

$T \ = \ \displaystyle\int_{\theta_1}^{\theta_2} \ \frac{1}{v(\theta)} \ L \ d\theta$

Integrating from $\frac{\pi}{2}$ to $0$ , we get one fourth of the period, thus:

$T \ = \ 4 \ \displaystyle\int_{0}^{\frac{\pi}{2}} \ \frac{L}{v(\theta)} \ d\theta \ = \ 4 \ \displaystyle\int_{0}^{\frac{\pi}{2}} \ \frac{L}{\sqrt{2gL \cos\ \theta}} \ d\theta \ = \ 2 \sqrt{\frac{2L}{g}} \ \displaystyle\int_{0}^{\frac{\pi}{2}} \ \frac{1}{\sqrt{\cos\ \theta}} \ = \ \sqrt{\frac{2L}{g}} \ \text{B}(\frac{1}{2}, \ \frac{1}{4})$

$\Rightarrow \ \sqrt{\frac{2L}{g}} \ \frac{\Gamma(\frac{1}{2}) \ \Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})} \ = \ \sqrt{\frac{L}{\pi g}} \ \Gamma(\frac{1}{4})^2$

So we have:

$\alpha \ + \ \beta \ + \ \gamma \ + \eta^{\delta} \ = \ 1\ + \ 1 \ + \ 1 \ + \ 4^2 \ = \ 19$