If I_m,n = _0^1 x^m-1(1-x)^n-1,dx, for m,n1, and _0^1x^m-1+x^n-1(1+x)^m+n,dx = ,I_m,n, R, …

Question

If $I_{m,n} = \displaystyle\int_0^1 x^{m-1}(1-x)^{n-1}\,dx$, for $m,n\ge1$, and $\displaystyle\int_0^1\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}\,dx = \alpha\,I_{m,n}$, $\alpha\in\mathbb{R}$, then $\alpha$ equals \_\_\_\_\_\_.

Answer: 1

Accepted Answer

Using the standard Beta-function identity $I_{m,n}=B(m,n)=\displaystyle\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\,dx$, split at $x=1$ and substitute $x=1/u$ on $(1,\infty)$:

$$\int_1^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\,dx = \int_0^1\frac{u^{n-1}}{(1+u)^{m+n}}\,du$$

So $I_{m,n} = \displaystyle\int_0^1\frac{x^{m-1}}{(1+x)^{m+n}}\,dx+\int_0^1\frac{x^{n-1}}{(1+x)^{m+n}}\,dx = \int_0^1\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}\,dx$

Hence $\alpha=1$.

Solutions · 1