Relation Between Beta and Gamma Functions

The Beta function $B(x, y)$ and the Gamma function $\Gamma(z)$ are connected through the following key relation:

$B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x + y)} \quad \text{where } x, y > 0.$

Proof of the Relation

Step 1: Definition of Beta Function

The Beta function is defined as: $B(x, y) = \int_0^1 t^{x-1} (1-t)^{y-1} \, dt, \quad x, y > 0.$

Step 2: Substituting $t = u/(1+u)$

Let us substitute $t = \frac{u}{1+u}$, which implies: $dt = \frac{1}{(1+u)^2} \, du.$ Also, $1-t = \frac{1}{1+u}$, and as $t$ goes from 0 to 1, $u$ goes from 0 to $\infty$.

Rewriting the Beta function under this substitution: $B(x, y) = \int_0^\infty \left(\frac{u}{1+u}\right)^{x-1} \left(\frac{1}{1+u}\right)^{y-1} \frac{1}{(1+u)^2} \, du.$

Simplify: $B(x, y) = \int_0^\infty u^{x-1} (1+u)^{-(x+y)} \, du.$

Step 3: Relating to the Gamma Function

The Gamma function is defined as: $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \, dt, \quad z > 0.$

Using the substitution $t = u$ and identifying $e^{-t}$ with $(1+u)^{-(x+y)}$, it turns out that the integral for $B(x, y)$ can be split into two Gamma functions.

By algebraic manipulation, this yields: $B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.$

Conclusion

Thus, the relation between the Beta and Gamma functions is proven: $B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.$

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Related Questions:

1. What are some real-world applications of the Beta and Gamma functions?
2. How does the Gamma function generalize the factorial function?
3. Can we derive the properties of the Beta function from its relation with the Gamma function?
4. How do we evaluate $B(x, y)$ for specific values of $x$ and $y$?
5. What is the significance of the substitution used in the proof?

Tip:

When solving integrals involving the Beta or Gamma function, carefully analyze the limits and substitution techniques to simplify computations.