Difference between revisions of "Integral of (1+bt^z)^(-y)t^x dt = (1/z)*b^(-(x+1)/z) B((x+1)/z,y-(x+1)/z)"
From specialfunctionswiki
(Created page with "==Theorem== The following formula holds for $z>0$, $b>0$, and $0 < \mathrm{Re} \left( \dfrac{x+1}{z} \right) < \mathrm{Re}(y)$: $$\displaystyle\int_0^{\infty} (1+bt^z)^{-y} t^...") |
|||
| Line 7: | Line 7: | ||
==References== | ==References== | ||
| − | * {{BookReference|Higher Transcendental Functions Volume I|1953| | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=Integral of (t-b)^(x-1)(a-t)^(y-1)/(c-t)^(x+y) dt = (a-b)^(x+y-1)/((c-a)^x (c-b)^y) B(x,y)|next=integral of t^(x-1)(1-t^z)^(y-1) dt=(1/z)B(x/z,y)}}: $\S 1.5 (16)$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 21:03, 3 March 2018
Theorem
The following formula holds for $z>0$, $b>0$, and $0 < \mathrm{Re} \left( \dfrac{x+1}{z} \right) < \mathrm{Re}(y)$: $$\displaystyle\int_0^{\infty} (1+bt^z)^{-y} t^x \mathrm{d}t = \dfrac{1}{z} b^{-\frac{x+1}{z}} B \left( \dfrac{x+1}{z}, y - \dfrac{x+1}{z} \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): $\S 1.5 (16)$
