Difference between revisions of "Beta in terms of sine and cosine"
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==Proof== | ==Proof== | ||
| + | From the definition, | ||
| + | $$B(x,y)=\displaystyle\int_0^1 \xi^{x-1} (1-\xi)^{y-1} \mathrm{d}\xi.$$ | ||
| + | Let $\xi=\sin^2(t)$. Then $d\xi = 2\sin(t)\cos(t)$. Also if $\xi=0$ then $0=\sin^2(t)$ implies that $t=\arcsin(0)=0$ and if $\xi=1$, then $1=\sin^2(t)$ implies $t=\arcsin(1)=\dfrac{\pi}{2}$. Therefore using [[substitution]] and the [[Pythagorean identity for sin and cos]], | ||
| + | $$\begin{array}{ll} | ||
| + | B(x,y) &= \displaystyle\int_0^1 \xi^{x-1}(1-\xi)^{y-1} \mathrm{d}\xi \\ | ||
| + | &= \displaystyle\int_0^{\frac{\pi}{2}} (\sin(t))^{2x-2} (1-\sin^2(t))^{y-1} 2 \sin(t)\cos(t) \mathrm{d}t \\ | ||
| + | &= 2 \displaystyle\int_0^{\frac{\pi}{2}} (\sin(t))^{2x-1} (\cos(t))^{2y-1} \mathrm{d}t, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. █ | ||
==References== | ==References== | ||
| + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=Integral of (1+t)^(2x-1)(1-t)^(2y-1)(1+t^2)^(-x-y)dt=2^(x+y-2)B(x,y)|next=findme}}: $\S 1.5 (19)$ | ||
| + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Beta in terms of power of t over power of (1+t)|next=Beta in terms of gamma}}: $6.2.1$ | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
| − | [[Category: | + | [[Category:Proven]] |
Latest revision as of 21:04, 3 March 2018
Theorem
The following formula holds: $$B(x,y)=2 \displaystyle\int_0^{\frac{\pi}{2}} (\sin t)^{2x-1}(\cos t)^{2y-1} \mathrm{d}t,$$ where $B$ denotes the beta function, $\sin$ denotes the sine function, and $\cos$ denotes the cosine function.
Proof
From the definition, $$B(x,y)=\displaystyle\int_0^1 \xi^{x-1} (1-\xi)^{y-1} \mathrm{d}\xi.$$ Let $\xi=\sin^2(t)$. Then $d\xi = 2\sin(t)\cos(t)$. Also if $\xi=0$ then $0=\sin^2(t)$ implies that $t=\arcsin(0)=0$ and if $\xi=1$, then $1=\sin^2(t)$ implies $t=\arcsin(1)=\dfrac{\pi}{2}$. Therefore using substitution and the Pythagorean identity for sin and cos, $$\begin{array}{ll} B(x,y) &= \displaystyle\int_0^1 \xi^{x-1}(1-\xi)^{y-1} \mathrm{d}\xi \\ &= \displaystyle\int_0^{\frac{\pi}{2}} (\sin(t))^{2x-2} (1-\sin^2(t))^{y-1} 2 \sin(t)\cos(t) \mathrm{d}t \\ &= 2 \displaystyle\int_0^{\frac{\pi}{2}} (\sin(t))^{2x-1} (\cos(t))^{2y-1} \mathrm{d}t, \end{array}$$ as was to be shown. █
References
- 1953: Arthur Erdélyi, Wilhelm Magnus, Fritz Oberhettinger and Francesco G. Tricomi: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (19)$
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $6.2.1$
