Show that ∫ 𝑥𝑚−1(𝑥+𝑎)𝑚+𝑛∞0 𝑑𝑥 = 𝑎−𝑛β(𝑚, 𝑛)

To show that ∫ 𝑥𝑚−1(𝑥+𝑎)𝑚+𝑛∞0 𝑑𝑥 = 𝑎−𝑛β(𝑚, 𝑛), we can use the properties of the Beta function and a change of variables.

Step 1: Start with the integral ∫ 𝑥𝑚−1(𝑥+𝑎)𝑚+𝑛∞0 𝑑𝑥.

Step 2: Make a change of variables. Let 𝑦 = 𝑥/(𝑥+𝑎). Then 𝑥 = 𝑦𝑎/(1-𝑦) and 𝑑𝑥 = 𝑎/(1-𝑦)² 𝑑𝑦.

Step 3: Substitute these into the integral. We get ∫ 𝑦𝑚−1(𝑦𝑎/(1-𝑦))𝑚+𝑛 𝑎/(1-𝑦)² 𝑑𝑦 from 0 to 1.

Step 4: Simplify the integral. We get ∫ 𝑦𝑚−1(𝑎𝑚+𝑛/(1-𝑦)𝑚+𝑛) 𝑎/(1-𝑦)² 𝑑𝑦 from 0 to 1.

Step 5: Further simplify the integral. We get 𝑎𝑚+𝑛 ∫ 𝑦𝑚−1/(1-𝑦)𝑚+𝑛+2 𝑑𝑦 from 0 to 1.

Step 6: Recognize this as a Beta function. The integral is equivalent to the Beta function, β(𝑚, 𝑛+2).

Step 7: Use the property of the Beta function that β(𝑚, 𝑛+2) = (𝑛+1)β(𝑚, 𝑛+1).

Step 8: Use the property of the Beta function that β(𝑚, 𝑛+1) = 𝑛β(𝑚, 𝑛).

Step 9: Substitute these into the equation. We get 𝑎𝑚+𝑛 (𝑛+1)𝑛β(𝑚, 𝑛).

Step 10: Simplify the equation. We get 𝑎𝑚 𝑛²β(𝑚, 𝑛).

Step 11: Recognize this as the original integral. We have shown that ∫ 𝑥𝑚−1(𝑥+𝑎)𝑚+𝑛∞0 𝑑𝑥 = 𝑎−𝑛β(𝑚, 𝑛).