To prove that $\beta(m, n) = \beta(n, m)$, let us analyze what the Beta function represents. The Beta function is defined as:

$\beta(m, n) = \int_0^1 t^{m-1}(1-t)^{n-1} \, dt$

Proof:

We start with the definition: $\beta(m, n) = \int_0^1 t^{m-1}(1-t)^{n-1} \, dt$

Now, make the substitution $u = 1 - t$. Then:

• $du = -dt$
• When $t = 0$, $u = 1$
• When $t = 1$, $u = 0$

Substituting into the integral: $\beta(m, n) = \int_1^0 (1-u)^{m-1} u^{n-1} (-du)$

Simplify: $\beta(m, n) = \int_0^1 u^{n-1} (1-u)^{m-1} \, du$

Notice that this is exactly the definition of $\beta(n, m)$: $\beta(n, m) = \int_0^1 u^{n-1}(1-u)^{m-1} \, du$

Thus: $\beta(m, n) = \beta(n, m)$

Conclusion:

We have shown that the Beta function is symmetric, meaning $\beta(m, n) = \beta(n, m)$.

Would you like a more detailed explanation of any part of this proof? Let me know!

Related Questions:

1. What is the relationship between the Beta function and the Gamma function?
2. How does symmetry in the Beta function help in evaluating integrals?
3. Can the Beta function be extended to complex arguments?
4. How is the Beta function used in probability and statistics?
5. What is the geometric interpretation of the Beta function?

Tip:

For solving problems with the Beta function, always check for symmetry—it can simplify calculations significantly!