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I am aware that a proof for the Euler's Beta function for positive integers has already been asked, but I have tried to derive it myself without using the gamma function, and my result doesn't match. Any ideas?

My derivation (I used $n=sm+d$ and $t=\left(1-x\right)^m$): \begin{align*} B\left(m,n\right)&=\int_{0}^{1}{x^{m-1}\left(1-x\right)^{n-1}dx} \\ &=\frac{1}{m}\int_{0}^{1}{\left(1-x\right)^{n-1}dx^m} \\ &=\frac{1}{m}\int_{1}^{0}{\left(1-x\right)^{n-1}d\left(1-x\right)^m} \\ &=-\frac{1}{m}\int_{0}^{1}{\left(1-x\right)^{sm+d-1}d\left(1-x\right)^m} \\ &=-\frac{1}{m}\int_{1}^{0}{t^{\frac{sm+d-1}{m}}dt} \\ &=\frac{1}{m}\int_{0}^{1}{t^{\frac{sm+d-1}{m}}dt} \\ &=\frac{1}{m}\left(\frac{m}{sm+m+d}1^{\frac{sm+m+d-1}{m}}-\frac{m}{sm+m+d}0^{\frac{sm+m+d-1}{m}}\right) \\ &=\frac{1}{sm+m+d-1} \\ &=\frac{1}{m+n-1} \\ \end{align*}

Correct result:

$$B\left(m,n\right)=\frac{\left(m−1\right)!\cdot\left(n−1\right)!}{\left(m+n−1\right)!}$$

Tiger Yu
  • 111

1 Answers1

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Ok, I've found a solution:

let $a_i=\int_{0}^{1}{\left(m+i-1\right)x^{m+i-2}\left(1-x\right)^{n-i}dx}$

let $u_i=\left(1-x\right)^{n-i},du_i=\left(i-n\right)\left(1-x\right)^{n-i-1}dx$

let $v_i=x^{m+i-1},dv_i=\left(m+i-1\right)x^{m+i-2}dx$

\begin{align*} a_i&=\int_{0}^{1}{\left(m+i-1\right)x^{m+i-2}\left(1-x\right)^{n-i}dx}\\ &=\int_{0}^{1}{u_idv_i}\\ &=\left[u_iv_i\right]_0^1+\int_{0}^{1}{v_idu_i}\\ &=\left[\left(1-v_i^{\frac{1}{m+i-1}}\right)^{n-i}v_i\right]_0^1-\int_{0}^{1}{x^{m+i-1}\left(i-n\right)\left(1-x\right)^{n-i-1}dx}\\ &=0-\left(i-n\right)\int_{0}^{1}{x^{m+i-1}\left(1-x\right)^{n-i-1}dx}\\ &=\frac{n-i}{m+i}\int_{0}^{1}{\left(m+i\right)x^{m+i-1}\left(1-x\right)^{n-i-1}dx}\\ &=\frac{n-i}{m+i}a_{i+1}\\ &=a_{n-1}\prod_{j=1}^{n-2}{\frac{n-j}{m+j}}\\ &=\frac{m!\left(n-1\right)!}{\left(m+n-2\right)!}a_{n-1} \end{align*}

\begin{align*} B\left(m,n\right)&=\int_{0}^{1}{x^{m-1}\left(1-x\right)^{n-1}dx}\\ &=\frac{1}{m}\int_{0}^{1}{mx^{m-1}\left(1-x\right)^{n-1}dx}\\ &=\frac{1}{m}a_1\\ &=\frac{1}{m}\frac{m!\left(n-1\right)!}{\left(m+n-2\right)!}a_{n-1}\\ &=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\int_{0}^{1}{\left(m+n-2\right)x^{m+n-3}\left(1-x\right)dx}\\ &=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left[\int_{0}^{1}{\left(m+n-2\right)x^{m+n-3}dx}-\int_{0}^{1}{\left(m+n-2\right)x^{m+n-2}dx}\right]\\ &=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left(\left[x^{m+n-2}\right]_0^1-\left[\frac{m+n-2}{m+n-1}x^{m+n+1}\right]_0^1\right)\\ &=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left(1-\frac{m+n-2}{m+n-1}\right)\\ &=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}-\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\frac{m+n-2}{m+n-1}\\ &=\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-1\right)}{\left(m+n-2\right)!}-\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-2\right)}{\left(m+n-1\right)!}\\ &=\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-1-m-n+2\right)}{\left(m+n-2\right)!}\\ &=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-1\right)!} \end{align*}

Tiger Yu
  • 111