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Beta integral I(m, n): identities and evaluation

This type of question, asked 3 times: 2021 Paper 26, 5 Apr 2024, 24 Jan 2025. All 3 questions below, oldest first.

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  1. Question 1

    2021 Paper 26

    84
    In Im,n=∫01xm−1(1−x)n−1dxI_{m,n}=\int_{0}^{1} x^{m - 1}(1 - x)^{n - 1}dx, for m,n≥1m,n \geq 1 and ∫01xm−1+xn−1(1+x)m+ndx=αIm,n\int_{0}^{1} \frac{x^{m - 1}+x^{n - 1}}{(1 + x)^{m + n}}dx = \alpha I_{m,n}, α∈R\alpha \in R, then α\alpha equals

  2. Question 2

    5 Apr 2024

    167
    Let β(m,n)=∫01xm−1(1−x)n−1dx,m,n>0\beta(m,n) =\int_{0}^{1} x^{m - 1}(1 - x)^{n - 1}dx,m,n > 0. If ∫01(1−x10)20dx=a×β(b,c)\int_{0}^{1} \left( 1 -x^{10} \right)^{20}dx = a \times \beta(b,c), then 100(a+b+c)100(a + b + c) equals

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  3. Question 3

    24 Jan 2025

    52
    In I(m,n)=∫01xm−1(1−x)n−1dx,m,n>0I(m,n) =\int_{0}^{1} x^{m - 1}(1 - x)^{n - 1}dx,m,n > 0, then I(9,14)+I(10,13)I(9,14) + I(10,13) is

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