Difference between revisions of "Integral of t^(x-1)(1-t^z)^(y-1) dt=(1/z)B(x/z,y)"
From specialfunctionswiki
(Created page with "==Theorem== The following formula holds for $z>0$, $\mathrm{Re}(y)>0$, and $\mathrm{Re}(x)>0$: $$\displaystyle\int_0^1 t^{x-1} (1-t^z)^{y-1} \mathrm{d}t = \dfrac{1}{z} B \left...") |
|||
| Line 7: | Line 7: | ||
==References== | ==References== | ||
| − | * {{BookReference|Higher Transcendental Functions Volume I|1953| | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=Integral of (1+bt^z)^(-y)t^x dt = (1/z)*b^(-(x+1)/z) B((x+1)/z,y-(x+1)/z)|next=Integral of (1+t)^(2x-1)(1-t)^(2y-1)(1+t^2)^(-x-y)dt=2^(x+y-2)B(x,y)}}: $\S 1.5 (17)$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 21:03, 3 March 2018
Theorem
The following formula holds for $z>0$, $\mathrm{Re}(y)>0$, and $\mathrm{Re}(x)>0$: $$\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.
Proof
References
- 1953: Arthur Erdélyi, Wilhelm Magnus, Fritz Oberhettinger and Francesco G. Tricomi: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (17)$
