Difference between revisions of "Integral of (t-b)^(x-1)(a-t)^(y-1)/(t-x)^(x+y) dt=(a-b)^(x+y-1)/((a-c)^x(b-c)^y) B(x,y)"
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(Created page with "==Theorem== The following formula holds for $\mathrm{Re}(x)>0$, $\mathrm{Re}(y) > 0$, and $c<b<a$: $$\displaystyle\int_a^b \dfrac{(t-b)^{x-1}(a-t)^{y-1}}{(t-c)^{x+y}} \mathrm{...") |
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==References== | ==References== | ||
| − | * {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=Integral (t-b)^(x-1)(a-t)^(y-1)dt=(a-b)^(x+y-1)B(x,y)|next= | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=Integral (t-b)^(x-1)(a-t)^(y-1)dt=(a-b)^(x+y-1)B(x,y)|next=integral of (t-b)^(x-1)(a-t)^(y-1)/(c-t)^(x+y) dt = (a-b)^(x+y-1)/((c-a)^x (c-b)^y) B(x,y)}}: $\S 1.5 (14)$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Revision as of 23:18, 24 June 2017
Theorem
The following formula holds for $\mathrm{Re}(x)>0$, $\mathrm{Re}(y) > 0$, and $c<b<a$: $$\displaystyle\int_a^b \dfrac{(t-b)^{x-1}(a-t)^{y-1}}{(t-c)^{x+y}} \mathrm{d}t=\dfrac{(a-b)^{x+y-1}}{(a-c)^x (b-c)^y} B(x,y),$$ where $B$ denotes the beta function.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (14)$
