View source for Integral of (1+bt^z)^(-y)t^x dt = (1/z)*b^(-(x+1)/z) B((x+1)/z,y-(x+1)/z)

==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==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|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:Unproven]]