$int_0^xsin^atcos^bt, mathrm{d}t=?$
Consider the integral
$$I(a,b)=int_0^{pi/2}sin^at cos^bt dt$$
As I have shown in numerous answers, $I$ has a close relationship with the beta function, namely
$$I(a,b)=frac12Bbigg(frac{a+1}2,frac{b+1}2bigg)$$
Which of course lends itself to a representation via the gamma function
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
And similarly,
$$int_0^xsin^at cos^bt dt=frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)$$
Where $B(x;a,b)$ is the Incomplete Beta function.
But here's my question:
Let $$G_{ab}(x)=frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)$$
Assuming $a,b>-1$ are held constant, what is the relationship between $G_{ab}(frac{pi}2)$ and $G_{ab}(x)$ at other values of $xgeq0$. I don't expect that there is a closed form expression for all $x$, but I'd expect there'd be one for $$xinbigg{frac{pi}2(2n+1):ninBbb Nbigg}$$
As $sin^2x=1$ for such $x$.
My instinct tells me that for such $x$, $$G_{ab}(x)=I(a,b)$$
But I just don't know how to prove it. Could I have some help?
integration special-functions trigonometric-integrals
asked Nov 14 '18 at 18:26
clathratusclathratus
3,422331
add a comment |
Consider the integral
$$I(a,b)=int_0^{pi/2}sin^at cos^bt dt$$
As I have shown in numerous answers, $I$ has a close relationship with the beta function, namely
$$I(a,b)=frac12Bbigg(frac{a+1}2,frac{b+1}2bigg)$$
Which of course lends itself to a representation via the gamma function
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
And similarly,
$$int_0^xsin^at cos^bt dt=frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)$$
Where $B(x;a,b)$ is the Incomplete Beta function.
But here's my question:
Let $$G_{ab}(x)=frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)$$
Assuming $a,b>-1$ are held constant, what is the relationship between $G_{ab}(frac{pi}2)$ and $G_{ab}(x)$ at other values of $xgeq0$. I don't expect that there is a closed form expression for all $x$, but I'd expect there'd be one for $$xinbigg{frac{pi}2(2n+1):ninBbb Nbigg}$$
As $sin^2x=1$ for such $x$.
My instinct tells me that for such $x$, $$G_{ab}(x)=I(a,b)$$
But I just don't know how to prove it. Could I have some help?
integration special-functions trigonometric-integrals
asked Nov 14 '18 at 18:26
clathratusclathratus
3,422331
Just to clarity it's $G_{ab}$ not $G_a,b$ ?
– DavidG
Dec 9 '18 at 11:56
@DavidG Yes. I guess we could also write it as $$G_{a,b}$$
– clathratus
Dec 9 '18 at 18:33
add a comment |
Consider the integral
$$I(a,b)=int_0^{pi/2}sin^at cos^bt dt$$
As I have shown in numerous answers, $I$ has a close relationship with the beta function, namely
$$I(a,b)=frac12Bbigg(frac{a+1}2,frac{b+1}2bigg)$$
Which of course lends itself to a representation via the gamma function
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
And similarly,
$$int_0^xsin^at cos^bt dt=frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)$$
Where $B(x;a,b)$ is the Incomplete Beta function.
But here's my question:
Let $$G_{ab}(x)=frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)$$
Assuming $a,b>-1$ are held constant, what is the relationship between $G_{ab}(frac{pi}2)$ and $G_{ab}(x)$ at other values of $xgeq0$. I don't expect that there is a closed form expression for all $x$, but I'd expect there'd be one for $$xinbigg{frac{pi}2(2n+1):ninBbb Nbigg}$$
As $sin^2x=1$ for such $x$.
My instinct tells me that for such $x$, $$G_{ab}(x)=I(a,b)$$
But I just don't know how to prove it. Could I have some help?
integration special-functions trigonometric-integrals
asked Nov 14 '18 at 18:26
clathratusclathratus
3,422331
Consider the integral
$$I(a,b)=int_0^{pi/2}sin^at cos^bt dt$$
As I have shown in numerous answers, $I$ has a close relationship with the beta function, namely
$$I(a,b)=frac12Bbigg(frac{a+1}2,frac{b+1}2bigg)$$
Which of course lends itself to a representation via the gamma function
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
And similarly,
$$int_0^xsin^at cos^bt dt=frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)$$
Where $B(x;a,b)$ is the Incomplete Beta function.
But here's my question:
Let $$G_{ab}(x)=frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)$$
Assuming $a,b>-1$ are held constant, what is the relationship between $G_{ab}(frac{pi}2)$ and $G_{ab}(x)$ at other values of $xgeq0$. I don't expect that there is a closed form expression for all $x$, but I'd expect there'd be one for $$xinbigg{frac{pi}2(2n+1):ninBbb Nbigg}$$
As $sin^2x=1$ for such $x$.
My instinct tells me that for such $x$, $$G_{ab}(x)=I(a,b)$$
But I just don't know how to prove it. Could I have some help?
integration special-functions trigonometric-integrals
integration special-functions trigonometric-integrals
asked Nov 14 '18 at 18:26
clathratusclathratus
3,422331
asked Nov 14 '18 at 18:26
clathratusclathratus
3,422331
edited Dec 1 '18 at 5:02
clathratus
asked Nov 14 '18 at 18:26
clathratusclathratus
3,422331
asked Nov 14 '18 at 18:26
clathratusclathratus
3,422331
asked Nov 14 '18 at 18:26
clathratusclathratus
3,422331
3,422331
Just to clarity it's $G_{ab}$ not $G_a,b$ ?
– DavidG
Dec 9 '18 at 11:56
@DavidG Yes. I guess we could also write it as $$G_{a,b}$$
– clathratus
Dec 9 '18 at 18:33
add a comment |
Just to clarity it's $G_{ab}$ not $G_a,b$ ?
– DavidG
Dec 9 '18 at 11:56
@DavidG Yes. I guess we could also write it as $$G_{a,b}$$
– clathratus
Dec 9 '18 at 18:33
Just to clarity it's $G_{ab}$ not $G_a,b$ ?
– DavidG
Dec 9 '18 at 11:56
Just to clarity it's $G_{ab}$ not $G_a,b$ ?
– DavidG
Dec 9 '18 at 11:56
@DavidG Yes. I guess we could also write it as $$G_{a,b}$$
– clathratus
Dec 9 '18 at 18:33
@DavidG Yes. I guess we could also write it as $$G_{a,b}$$
– clathratus
Dec 9 '18 at 18:33
add a comment |
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Just to clarity it's $G_{ab}$ not $G_a,b$ ?
– DavidG
Dec 9 '18 at 11:56
@DavidG Yes. I guess we could also write it as $$G_{a,b}$$
– clathratus
Dec 9 '18 at 18:33