$I_k=\int_0^1 \frac{1}{\mathbf{B}(\alpha , \beta )} \cos^k (\pi \theta) \theta^{\alpha -1} (1-\theta)^{\beta -1}d\theta $

Question

dejar $\alpha>0$ , $\beta>0$ , $k\in \{1,2,\cdots \}$ y $\mathbf{B}(\alpha , \beta )=\frac{\Gamma (\alpha ) \Gamma (\beta)}{\Gamma (\alpha + \beta )}$ .

P1: ¿Cómo podemos resolver las siguientes integrales? $$I_k(\alpha , \beta)=\int_0^1 \frac{1}{\mathbf{B}(\alpha , \beta )} \cos^k (\pi \theta) \theta^{\alpha -1} (1-\theta)^{\beta -1}d\theta $$

$$J_k(\alpha , \beta)=\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{\mathbf{B}(\alpha , \beta )} \sin^k (\pi \theta) \theta^{\alpha -1} (1-\theta)^{\beta -1}d\theta$$

Mi intento de $k=1$ (utilizando la serie de Taylor ):

\begin {eqnarray*} I_1( \alpha , \beta )&=& \int_ {0}^{1} \frac {1}{ \mathbf {B}( \alpha , \beta )} \cos ( \pi \theta ) \theta ^{ \alpha -1} (1- \theta )^{ \beta -1} d \theta \nonumber \\ &=& \frac {1}{ \mathbf {B}( \alpha , \beta )} \int_ {0}^{1} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n)!}( \pi \theta )^{2n} \theta ^{ \alpha -1} (1- \theta )^{ \beta -1} d \theta \nonumber \\ & \overset {DominatedCT}{=}& \frac {1}{ \mathbf {B}( \alpha , \beta )} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n)!}( \pi )^{2n} \int_ {0}^{1} \theta ^{2n+ \alpha -1} (1- \theta )^{ \beta -1} d \theta \nonumber \\ &=& \frac {1}{ \mathbf {B}( \alpha , \beta )} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n)!}( \pi )^{2n} \mathbf {B}(2n+ \alpha , \beta ) \\ &=& \frac { \Gamma ( \alpha + \beta )}{ \Gamma ( \alpha )} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n)!}( \pi )^{2n} \frac { \Gamma (2n+ \alpha )}{ \Gamma (2n+ \alpha + \beta )} \\ &=& 1+ \sum_ {n=1}^{ \infty } \frac {(-1)^n}{(2n)!}( \pi )^{2n} \frac {(2n+ \alpha -1 )(2n+ \alpha -2) \cdots ( \alpha )}{(2n+ \alpha + \beta -1 )(2n+ \alpha + \beta -2 ) \cdots ( \alpha + \beta ) } \\ &=& 1+ \sum_ {n=1}^{ \infty } \frac {(-1)^n}{(2n)!}( \pi )^{2n} \prod_ {r=0}^{2n-1} \frac { \alpha +r}{ \alpha + \beta +r} \end {eqnarray*}

Puedo utilizar la serie Taylor de $\cos^k(\pi \theta)$ pero quiero una forma cerrada para $I_1(\alpha , \beta)$ y $I_2(\alpha , \beta)$ . es decir, necesito resolver la suma anterior.

P2: ¿Existe una forma mejor?

He buscado un poco pero no he conseguido nada.

\begin {eqnarray} I_2( \alpha , \beta )&=& \int_ {0}^{1} \frac {1}{ \mathbf {B}( \alpha , \beta )} \cos ^2 ( \pi \theta ) \theta ^{ \alpha -1} (1- \theta )^{ \beta -1} d \theta \\ &=& \frac {1}{ \mathbf {B}( \alpha , \beta )} \int_ {0}^{1} \left ( 1+ \sum_ {n=1}^{ \infty } \frac {(-1)^n}{2(2n)!}(2 \pi \theta )^{2n} \right ) \theta ^{ \alpha -1} (1- \theta )^{ \beta -1} d \theta \\ &=& 1+ \frac {1}{2 \mathbf {B}( \alpha , \beta )} \int_ {0}^{1} \sum_ {n=1}^{ \infty } \frac {(-1)^n}{(2n)!}(2 \pi )^{2n} \int_ {0}^{1} \theta ^{2n+ \alpha -1} (1- \theta )^{ \beta -1} d \theta \\ &=& 1+ \frac {1}{2 \mathbf {B}( \alpha , \beta )} \sum_ {n=1}^{ \infty } \frac {(-1)^n}{(2n)!}(2 \pi )^{2n} \mathbf {B}(2n+ \alpha , \beta ) \\ &=& 1+ \frac {1}{2} \sum_ {n=1}^{ \infty } \frac {(-1)^n}{(2n)!}(2 \pi )^{2n} \prod_ {r=0}^{2n-1} \frac { \alpha +r}{ \alpha + \beta +r} \end {eqnarray}

He encontrado una forma mejor de calcular $I_k(\alpha , \beta )$ y $J_k(\alpha , \beta )$ reduciendo la potencia.

definir

\begin {eqnarray} I^{(m)}( \alpha , \beta )&=& \int_0 ^{1} \frac {1}{ \mathbf {B}( \alpha , \beta )} \cos (m \pi \theta ) \theta ^{ \alpha -1} (1- \theta )^{ \beta -1} d \theta \nonumber \\ &=& \frac {1}{ \mathbf {B}( \alpha , \beta )} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n)!}(m \pi )^{2n} \int_ {0}^{1} \theta ^{2n+ \alpha -1} (1- \theta )^{ \beta -1} d \theta \nonumber \\ &=& \frac {1}{ \mathbf {B}( \alpha , \beta )} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n)!}(m \pi )^{2n} \mathbf {B}(2n+ \alpha , \beta ) \nonumber \\ &=& \frac { \Gamma ( \alpha + \beta )}{ \Gamma ( \alpha )} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n)!}(m \pi )^{2n} \frac { \Gamma (2n+ \alpha )}{ \Gamma (2n+ \alpha + \beta )} \nonumber \\ &=& 1+ \sum_ {n=1}^{ \infty } \frac {(-1)^n}{(2n)!}(m \pi )^{2n} \frac {(2n+ \alpha -1 )(2n+ \alpha -2) \cdots ( \alpha )}{(2n+ \alpha + \beta -1 )(2n+ \alpha + \beta -2 ) \cdots ( \alpha + \beta ) } \nonumber \\ &=& 1+ \sum_ {n=1}^{ \infty } \frac {(-1)^n}{(2n)!}(m \pi )^{2n} \prod_ {r=0}^{2n-1} \frac { \alpha +r}{ \alpha + \beta +r} \end {eqnarray}

\begin {eqnarray} J^{(m)}( \alpha , \beta )&=& \int_0 ^{1} \frac {1}{ \mathbf {B}( \alpha , \beta )} \sin (m \pi \theta ) \theta ^{ \alpha -1} (1- \theta )^{ \beta -1} d \theta \nonumber \\ &=& \int_0 ^{1} \frac {1}{ \mathbf {B}( \alpha , \beta )} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n+1)!}(m \pi )^{2n+1} \theta ^{2n+1 + \alpha -1} (1- \theta )^{ \beta -1} d \theta \nonumber \\ &=& \frac {1}{ \mathbf {B}( \alpha , \beta )} \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n+1)!}(m \pi )^{2n+1} \mathbf {B}(2n+1+ \alpha , \beta ) \nonumber \\ &=& \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n+1)!}(m \pi )^{2n+1} \frac { \mathbf {B}(2n+1+ \alpha , \beta )}{ \mathbf {B}( \alpha , \beta )} \nonumber \\ &=& \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n+1)!}(m \pi )^{2n+1} \frac { \Gamma (2n+1+ \alpha ) \Gamma ( \alpha + \beta )}{ \Gamma ( \alpha ) \Gamma (2n+1 + \alpha + \beta ) } \nonumber \\ &=& \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n+1)!}(m \pi )^{2n+1} \frac { (2n+ \alpha ) (2n+ \alpha -1) \cdot ( \alpha )}{ (2n + \alpha + \beta ) (2n + \alpha + \beta ) \cdots ( \alpha + \beta )} \nonumber \\ &=& \sum_ {n=0}^{ \infty } \frac {(-1)^n}{(2n+1)!}(m \pi )^{2n+1} \prod_ {r=0}^{2n} \frac { \alpha + r}{ \alpha + \beta +r} \nonumber \end {eqnarray}

por lo que al reducir la fórmula de potencia

\begin {eqnarray*} \cos ^2( \pi \theta )&=& \frac {1}{2} \left (1+ \cos (2 \pi \theta ) \right ) \\ \cos ^3( \pi \theta )&=& \frac {3}{4} \cos ( \pi \theta )+ \frac {1}{4} \cos (3 \pi \theta ) \\ \sin ^2( \pi \theta )&=& \frac {1}{2} \left (1- \cos (2 \pi \theta ) \right ) \\ \cos ^3( \pi \theta )&=& \frac {3}{4} \sin ( \pi \theta )+ \frac {1}{4} \sin (3 \pi \theta ) \end {eqnarray*}

así que

\begin {eqnarray*} I_2( \alpha , \beta )&=& \frac {1}{2} \left (1+ I^{(2)}( \alpha , \beta ) \right ) \\ I_3( \alpha , \beta )&=& \frac {3}{4} I^{(1)}( \alpha , \beta )+ \frac {1}{4} I^{(3)}( \alpha , \beta ) \\ J_2( \alpha , \beta )&=& \frac {1}{2} \left (1- I^{(2)}( \alpha , \beta ) \right ) \\ J_3( \alpha , \beta )&=& \frac {3}{4} J^{(1)}( \alpha , \beta )+ \frac {1}{4} J^{(3)}( \alpha , \beta ) \end {eqnarray*}

trabajar con $J^{(m)} (\alpha , \beta )$ y $J^{(m)} (\alpha , \beta )$ es mucho más fácil.

por lo que queremos la fórmula para $J^{(m)} (\alpha , \beta )$ y $J^{(m)} (\alpha , \beta )$ .

Answer 1

Añadido: forma de superíndice según Maple. Parece que estos funcionan incluso cuando $m$ no es un número entero. $$ I^{(m)}(a,b) = {}_2F_3\left( \frac{a}{2},\frac{a+1}{2}; \frac{1}{2},\frac{a+b}{2},\frac{a+b+1}{2}; \frac{-m^2\pi^2}{4}\right) \\ J^{(m)}(a,b) = \frac{m a \pi}{a+b}\;{}_2F_3\left( \frac{a+1}{2},\frac{a+2}{2}; \frac{3}{2},\frac{a+b+1}{2},\frac{a+b+2}{2}; \frac{-m^2\pi^2}{4} \right) $$