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)"
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(Created page with "==Theorem== The following formula holds for $z>0$, $b>0$, and $0 < \mathrm{Re} \left( \dfrac{x+1}{z} \right) < \mathrm{Re}(y)$: $$\displaystyle\int_0^{\infty} (1+bt^z)^{-y} t^...") |
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Revision as of 23:45, 24 June 2017
Theorem
The following formula holds for $z>0$, $b>0$, and $0 < \mathrm{Re} \left( \dfrac{x+1}{z} \right) < \mathrm{Re}(y)$: $$\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: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (16)$
