Integral of t^(x-1)(1-t^z)^(y-1) dt=(1/z)B(x/z,y)

Theorem

The following formula holds for $z>0$, $\mathrm{Re}(y)>0$, and $\mathrm{Re}(x)>0$: $$\displaystyle\int_0^1 t^{x-1} (1-t^z)^{y-1} \mathrm{d}t = \dfrac{1}{z} B \left( \dfrac{x}{z}, y \right),$$ where $B$ denotes the beta function.

References

• 1953: {{ #if: |{{{2}}}|Arthur Erdélyi}}{{#if: Wilhelm Magnus|{{#if: Fritz Oberhettinger|, {{ #if: |{{{2}}}|Wilhelm Magnus}}{{#if: Francesco G. Tricomi|, {{ #if: |{{{2}}}|Fritz Oberhettinger}}{{#if: |, {{ #if: |{{{2}}}|Francesco G. Tricomi}}{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and {{ #if: |{{{2}}}|Francesco G. Tricomi}}}}| and {{ #if: |{{{2}}}|Fritz Oberhettinger}}}}| and {{ #if: |{{{2}}}|Wilhelm Magnus}}}}|}}: [[Book:Arthur Erdélyi/Higher Transcendental Functions Volume I{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|Higher Transcendental Functions Volume I{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}}|{{#if: | ({{{ed}}} ed.)}}}}]]{{#if: | (translated by [[Mathematician:{{{translated}}}|{{ #if: |{{{2}}}|{{{translated}}}}}]])}}{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: Integral of (1+bt^z)^(-y)t^x dt = (1/z)*b^(-(x+1)/z) B((x+1)/z,y-(x+1)/z) | ... (previous)|}}{{#if: Integral of (1+t)^(2x-1)(1-t)^(2y-1)(1+t^2)^(-x-y)dt=2^(x+y-2)B(x,y) | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}: $\S 1.5 (17)$