If I_(m,n) = ∫₀¹ xᵐ⁻¹(1 - x)ⁿ⁻¹dx, for m, n ≥ 1, and
∫₀¹ (xᵐ⁻¹ + xⁿ⁻¹)/(1 + x)ᵐ⁺ⁿ dx = α I_(m,n)α ∈ R, then α equals ___________.

Solution

Maths sir · 4 steps

Put t = x/(1+x) to stretch I_(m,n) over 0 to ∞, then fold the range back at x = 1.

1. Change the variable. In I_(m,n) put t = x/(1+x). Then 1 - t = 1/(1+x) and dt = dx/(1+x)². As t goes from 0 to 1, x goes from 0 to ∞.

   I_(m,n) = ∫₀^∞ xᵐ⁻¹/(1+x)ᵐ⁻¹ · 1/(1+x)ⁿ⁻¹ · dx/(1+x)² = ∫₀^∞ xᵐ⁻¹/(1+x)ᵐ⁺ⁿ dx

2. Split at 1. Write ∫₀^∞ = ∫₀¹ + ∫₁^∞. In the second piece put x = 1/u, so dx = -du/u² and the limits 1 to ∞ become 1 to 0.
3. Fold it back. The integrand becomes (u⁻⁽ᵐ⁻¹⁾ uᵐ⁺ⁿ)/(1+u)ᵐ⁺ⁿ · 1/u² = uⁿ⁻¹/(1+u)ᵐ⁺ⁿ, and the minus sign flips the limits to 0 to 1. So ∫₁^∞ xᵐ⁻¹/(1+x)ᵐ⁺ⁿ dx = ∫₀¹ xⁿ⁻¹/(1+x)ᵐ⁺ⁿ dx.
4. So. Adding the two pieces, I_(m,n) = ∫₀¹ (xᵐ⁻¹ + xⁿ⁻¹)/(1+x)ᵐ⁺ⁿ dx, which is exactly the given integral. Hence α = 1.

The common slip: 2. Treating each term of the numerator as a full copy of I_(m,n) gives 2, not 1. Each term covers only one half of the range 0 to ∞, so together they make I_(m,n) once.

The idea to keep. The Beta integral has a second form: ∫₀¹ xᵐ⁻¹(1-x)ⁿ⁻¹dx = ∫₀^∞ xᵐ⁻¹/(1+x)ᵐ⁺ⁿdx. Any integral to ∞ can be folded onto 0 to 1 with x → 1/x.

Answer from the NTA final answer key. Solution by PrepWiser.