Difference between revisions of "Beta in terms of power of t over power of (1+t)"
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
| − | $$B(x,y)=\displaystyle\int_0^{\infty} \dfrac{t^{x-1}}{(1 | + | $$B(x,y)=\displaystyle\int_0^{\infty} \dfrac{t^{x-1}}{(t+1)^{x+y}},$$ |
where $B$ denotes the [[beta]] function. | where $B$ denotes the [[beta]] function. | ||
==Proof== | ==Proof== | ||
| + | From the definition, | ||
| + | $$B(x,y)=\displaystyle\int_0^1 u^{x-1} (1-u)^{y-1} \mathrm{d}u.$$ | ||
| + | We will proceed using [[substitution]]. Let $u=\dfrac{t}{t+1}$ so that $\mathrm{d}u=\dfrac{1}{(t+1)^2} \mathrm{d}t$. Since $u=0$ means $t=0$ and $u=1$ means $t=\infty$, we get | ||
| + | $$\begin{array}{ll} | ||
| + | B(x,y) &= \displaystyle\int_0^1 u^{x-1} (1-u)^{y-1} \mathrm{d}u \\ | ||
| + | &= \displaystyle\int_0^{\infty} \left( \dfrac{t}{t+1} \right)^{x-1} \left( 1 -\dfrac{t}{t+1} \right)^{y-1} \dfrac{1}{(t+1)^2} \mathrm{d}t \\ | ||
| + | &= \displaystyle\int_0^{\infty} \left( \dfrac{t}{t+1} \right)^{x-1} \left( \dfrac{1}{t+1} \right)^{y-1} \dfrac{1}{(t+1)^2} \mathrm{d}t \\ | ||
| + | &= \displaystyle\int_0^{\infty} \dfrac{t^{x-1}}{(t+1)^{x+y}}, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. | ||
==References== | ==References== | ||
| Line 10: | Line 20: | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
| − | [[Category: | + | [[Category:Proven]] |
Revision as of 15:25, 6 October 2016
Theorem
The following formula holds: $$B(x,y)=\displaystyle\int_0^{\infty} \dfrac{t^{x-1}}{(t+1)^{x+y}},$$ where $B$ denotes the beta function.
Proof
From the definition, $$B(x,y)=\displaystyle\int_0^1 u^{x-1} (1-u)^{y-1} \mathrm{d}u.$$ We will proceed using substitution. Let $u=\dfrac{t}{t+1}$ so that $\mathrm{d}u=\dfrac{1}{(t+1)^2} \mathrm{d}t$. Since $u=0$ means $t=0$ and $u=1$ means $t=\infty$, we get $$\begin{array}{ll} B(x,y) &= \displaystyle\int_0^1 u^{x-1} (1-u)^{y-1} \mathrm{d}u \\ &= \displaystyle\int_0^{\infty} \left( \dfrac{t}{t+1} \right)^{x-1} \left( 1 -\dfrac{t}{t+1} \right)^{y-1} \dfrac{1}{(t+1)^2} \mathrm{d}t \\ &= \displaystyle\int_0^{\infty} \left( \dfrac{t}{t+1} \right)^{x-1} \left( \dfrac{1}{t+1} \right)^{y-1} \dfrac{1}{(t+1)^2} \mathrm{d}t \\ &= \displaystyle\int_0^{\infty} \dfrac{t^{x-1}}{(t+1)^{x+y}}, \end{array}$$ as was to be shown.
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $6.2.1$
