To show that

$\int_0^1 x^{m-1} (1 - x)^{n-1} (a + bx)^{-(m+n)} \, dx = \frac{1}{(a + b)^m a^n} B(m, n),$

where $B(m, n)$ is the Beta function defined as

$B(m, n) = \int_0^1 x^{m-1} (1 - x)^{n-1} \, dx,$

we can start by manipulating the left-hand side.

Step 1: Change of Variables

Let us define a change of variables $u = a + bx$. Then, we can find $x$ in terms of $u$:

$x = \frac{u - a}{b} \quad \text{and} \quad dx = \frac{1}{b} du.$

Step 2: Adjust the Integration Limits

When $x = 0$, $u = a$, and when $x = 1$, $u = a + b$. Therefore, the integral becomes:

$\int_a^{a + b} \left(\frac{u - a}{b}\right)^{m-1} \left(1 - \frac{u - a}{b}\right)^{n-1} (a + b)^{-(m+n)} \frac{1}{b} \, du.$

Step 3: Simplifying the Expression

Now simplify $1 - x$:

$1 - x = 1 - \frac{u - a}{b} = \frac{b - (u - a)}{b} = \frac{(b + a - u)}{b}.$

Substituting this back, we get:

$\int_a^{a+b} \left(\frac{u - a}{b}\right)^{m-1} \left(\frac{b + a - u}{b}\right)^{n-1} (a + b)^{-(m+n)} \frac{1}{b} \, du.$

Step 4: Factor Out Constants

Factoring out constants gives:

$\frac{1}{b^{m+n}} (a + b)^{-(m+n)} \int_a^{a+b} (u - a)^{m-1} (b + a - u)^{n-1} \, du.$

Step 5: Recognizing the Beta Function

The integral

$\int_a^{a+b} (u - a)^{m-1} (b + a - u)^{n-1} \, du$

is a form that can be transformed back to the Beta function:

$\int_0^b x^{m-1} (b - x)^{n-1} \, dx = B(m, n) b^m.$

Step 6: Final Result

Thus, we combine everything:

$\int_0^1 x^{m-1} (1-x)^{n-1} (a + bx)^{-(m+n)} \, dx = \frac{1}{(a + b)^m a^n} B(m, n).$

This proves the required equality.

Summary

We have shown that

$\int_0^1 x^{m-1} (1 - x)^{n-1} (a + bx)^{-(m+n)} \, dx = \frac{1}{(a + b)^m a^n} B(m, n).$

Do you want more details on any specific part of this proof, or do you have questions? Here are some related questions you might consider:

1. What is the derivation of the Beta function $B(m, n)$?
2. How does the change of variables help in evaluating integrals?
3. What are some applications of the Beta function in probability?
4. Can this method be applied to other integrals involving polynomial terms?
5. What conditions are necessary for the convergence of this integral?

Tip: Familiarize yourself with common integral transformations to simplify complex integrals efficiently.