Relationship between Beta Function and Gamma Function
Theorem
$$ B(p,q) = {{\Gamma (p) \Gamma (q)} \over {\Gamma (p+q) }} $$
Explanation
The Beta function is defined as $\displaystyle B(p,q) := \int_{0}^{1} t^{p-1} (1-t)^{q-1} dt $, and, like the Gamma function, it is an important function applied in many fields. Since the Gamma function can be easily calculated using the recursive relationship, the Beta function can also be calculated easily using the above relation. Intuitively, it can be seen as a generalization of the binomial coefficient, and since factorials appear, it naturally has a lot of relations with the Gamma function.
Proof
If $\displaystyle \Gamma (p) = \int_{0}^{\infty} u^{p-1} e^{-u} du $ and $\displaystyle \Gamma (q) = \int_{0}^{\infty} v^{p-1} e^{-v} dv $, then $$ \begin{align*} \Gamma (p) \Gamma (q) &= \int_{0}^{\infty} u^{p-1} e^{-u} du \int_{0}^{\infty} v^{p-1} e^{-v} dv \\ &= \int_{0}^{\infty} \int_{0}^{\infty} u^{p-1} v^{q-1} e^{-u} e^{-v} du dv \end{align*} $$ Substituting with $u + v = z$, $u = zt$, and $v = z( 1 - t)$ to get $$ \begin{align*} \Gamma (p) \Gamma (q) &= \int_{0}^{\infty} \int_{0}^{1} (zt)^{p-1} (z(1-t))^{q-1} e^{-u-v} z dt dz \\ &= \int_{0}^{\infty} \int_{0}^{1} z^{p+q-1} t^{p-1} (1-t)^{q-1} e^{-z} dt dz \\ &= \int_{0}^{\infty} z^{p+q-1} e^{-z} \int_{0}^{1} t^{p-1} (1-t)^{q-1} dt dz \\ &= \int_{0}^{\infty} z^{p+q-1} e^{-z} dz \int_{0}^{1} t^{p-1} (1-t)^{q-1} dt \\ &= \Gamma (p + q) B(p, q) \end{align*} $$ Upon arranging, $$ {{\Gamma (p) \Gamma (q)} \over {\Gamma (p+q) }} = B(p,q) $$
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Corollary: Symmetry of the Beta Function
$$ B(p,q) = B(q,p) $$
It’s also simple to the extent of being foolish to prove it with substitution, as $\displaystyle B(p,q) = {{\Gamma (p) \Gamma (q)} \over {\Gamma (p+q) }} = {{\Gamma (q) \Gamma (p)} \over {\Gamma (q+p) }} = B(q,p) $ is enough.