Difference between revisions of "Beta in terms of sine and cosine"

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==Proof==
 
==Proof==
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From the definition,
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$$B(x,y)=\displaystyle\int_0^1 \xi^{x-1} (1-\xi)^{y-1} \mathrm{d}\xi.$$
 +
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]],
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$$\begin{array}{ll}
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B(x,y) &= \displaystyle\int_0^1 \xi^{x-1}(1-\xi)^{y-1} \mathrm{d}\xi \\
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&= \displaystyle\int_0^{\frac{\pi}{2}} (\sin(t))^{2x-2} (1-\sin^2(t))^{y-1} 2 \sin(t)\cos(t) \mathrm{d}t \\
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&= 2 \displaystyle\int_0^{\frac{\pi}{2}} (\sin(t))^{2x-1} (\cos(t))^{2y-1} \mathrm{d}t,
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\end{array}$$
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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}}: $\S 1.5 (19)$
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* {{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}}: $6.2.1$
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[[Category:Theorem]]
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[[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

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