Difference between revisions of "Integral t^(x-1)(1+bt)^(-x-y) dt = b^(-x) B(x,y)"
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(Created page with "==Theorem== The following formula holds for $b>0$, $\mathrm{Re}(x)>0$, and $\mathrm{Re}(y)>0$: $$\displaystyle\int_0^{\infty} t^{-x-1} (1+bt)^{-x-y} \mathrm{d}t=b^{-x}B(x,y),$...") |
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==References== | ==References== | ||
− | * {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev= | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=Integral t^(x-1)(1-t)^(y-1)(1+bt)^(-x-y)dt = (1+b)^(-x)B(x,y)|next=integral (t-b)^(x-1)(a-t)^(y-1)dt=(a-b)^(x+y-1)B(x,y)}}: $\S 1.5 (12)$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] |
Revision as of 23:56, 24 June 2017
Theorem
The following formula holds for $b>0$, $\mathrm{Re}(x)>0$, and $\mathrm{Re}(y)>0$: $$\displaystyle\int_0^{\infty} t^{-x-1} (1+bt)^{-x-y} \mathrm{d}t=b^{-x}B(x,y),$$ where $B$ denotes the beta function.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (12)$