State and prove relation between beta and gamma functions

To state and prove the relation between the beta and gamma functions, we start by defining the beta function as:

B(x, y) = ∫[0, 1] t^(x-1) * (1-t)^(y-1) dt

where x and y are positive real numbers.

Next, we define the gamma function as:

Γ(z) = ∫[0, ∞] t^(z-1) * e^(-t) dt

where z is a positive real number.

The relation between the beta and gamma functions is given by:

B(x, y) = Γ(x) * Γ(y) / Γ(x + y)

To prove this relation, we can use the properties of the gamma function.

First, we rewrite the beta function as:

B(x, y) = ∫[0, 1] t^(x-1) * (1-t)^(y-1) dt = ∫[0, ∞] (t / (1+t))^(x-1) * (1 / (1+t))^y * (1 / (1+t)) dt

Next, we make a substitution by letting u = t / (1+t). This gives us:

B(x, y) = ∫[0, 1] u^(x-1) * (1-u)^(y-1) * (1 / (1-u)^2) du

Now, we can rewrite the integral in terms of the gamma function:

B(x, y) = ∫[0, 1] u^(x-1) * (1-u)^(y-1) * (1 / (1-u)^2) du = ∫[0, 1] u^(x-1) * (1-u)^(y-1) * (1 / Γ(2)) * (Γ(2) / (1-u)^2) du = (1 / Γ(2)) * ∫[0, 1] u^(x-1) * (1-u)^(y-1) * (Γ(2) / (1-u)^2) du

Using the definition of the gamma function, we can simplify further:

B(x, y) = (1 / Γ(2)) * ∫[0, 1] u^(x-1) * (1-u)^(y-1) * (Γ(2) / (1-u)^2) du = (1 / Γ(2)) * ∫[0, 1] u^(x-1) * (1-u)^(y-1) * Γ(2) * (1-u)^(-2) du = (1 / Γ(2)) * Γ(x) * Γ(y) * ∫[0, 1] u^(x-1) * (1-u)^(y-1) * (1-u)^(-2) du = (Γ(x) * Γ(y)) / Γ(x + y)

Therefore, we have proved the relation between the beta and gamma functions:

B(x, y) = Γ(x) * Γ(y) / Γ(x + y)