Cómo podría probar esto: $F(x)=\int_{0}^{\pi/2} \cos^{x}\! \theta \cos(x\theta){\mathrm{d}}\theta$ = $\frac{\pi}{2^{a+1}}, \quad$

Question

Para $a=x>-1$

probarlo:

$F(x)=\int_{0}^{\pi/2} \cos^{x}\! \theta \cos(x\theta){\mathrm{d}}\theta$ = $\frac{\pi}{2^{a+1}}, \quad$

Me interesaría recibir cualquier respuesta o comentario

Answer 1

Utilicé esta fórmula que Evaluado por Augustin Louis Cauchy:

para $a>-1$ y $0 \leq b<a+2$ :

$\int_{0}^{\pi/2} \cos^{a}\! \theta \cos(b\theta) {\mathrm{d}}\theta = \frac{\pi}{2^{a+1}}\frac{\Gamma(a+1)}{\Gamma(1+\frac{a+b}{2})\Gamma(1+\frac{a-b}{2})}$

me parece que para a=b como caso especial obtengo :

$\int_{0}^{\pi/2} \cos^{a}\! \theta \cos(a\theta) {\mathrm{d}}\theta = \frac{\pi}{2^{a+1}}, \quad a>-1$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi/2}\cos^{a}\pars{\theta}\cos\pars{b\theta}\,\dd\theta ={\pi \over 2^{a+1}}:\ {\large ?}}$ .

$\ds{\large\tt\mbox{I'll consider the case}\quad a\,, b \in {\mathbb R}}$ :

$$\begin{align}&\color{#c00000}{\int_{0}^{\pi/2} \cos^{a}\pars{\theta}\cos\pars{b\theta}\,\dd\theta} =\Re\int_{0}^{\pi/2}\cos^{a}\pars{\theta}\expo{\ic b\theta}\,\dd\theta \\[3mm]&=\Re \int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi/2}} \pars{z^{2} + 1 \over 2z}^{a}z^{b}\,{\dd z \over \ic z} \\[3mm]&={1 \over 2^{a}}\,\Im\bracks{\color{#00f}{% \int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi/2}} \pars{z^{2} + 1}^{a}z^{b - a - 1}\,\dd z}}\tag{1} \end{align}$$

$\ds{\mbox{With}\ 0 < \epsilon < 1}$ : $$\begin{align}&\color{#00f}{% \int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi/2}} \pars{z^{2} + 1}^{a}z^{b - a - 1}\,\dd z} \\[3mm]&=-\int_{1}^{\epsilon}\pars{-y^{2} + 1}^{a}y^{b - a - 1} \expo{\ic\pars{b - a - 1}\pi/2}\ic\,\dd y \\[3mm]&\phantom{=} -\int_{\pi/2}^{0}\pars{\epsilon^{2}\expo{2\ic\theta} + 1}^{a}\epsilon^{b - a - 1} \expo{\ic\pars{b - a - 1}\theta}\epsilon\expo{\ic\theta}\ic\,\dd\theta -\int_{\epsilon}^{1}\pars{x^{2} + 1}^{a}x^{b - a - 1}\,\dd x \\[3mm]&=\half\,\expo{\ic\pars{b - a}\pi/2}\int_{\epsilon^{2}}^{1}\pars{1 - y}^{a} y^{\pars{b - a}/2 - 1}\,\dd y +\ic\epsilon^{b - a}\int_{0}^{\pi/2}\pars{\epsilon^{2}\expo{2\ic\theta} + 1}^{a} \expo{\ic\pars{b - a}\theta}\,\dd\theta \\[3mm]&\phantom{=}-\int_{\epsilon}^{1}\pars{x^{2} + 1}^{a}x^{b - a - 1}\,\dd x \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pars{2} \end{align}$$

En $\ds{\epsilon \to 0^{+}}$ Encuentro $\ds{\pars{~\mbox{from}\ \pars{1}\ \mbox{and}\ \pars{2}~}}$ : $$\begin{align}&\color{#c00000}{\int_{0}^{\pi/2} \cos^{a}\pars{\theta}\cos\pars{b\theta}\,\dd\theta} ={1 \over 2^{a + 1}}\,\sin\pars{\bracks{b - a}\pi \over 2} \int_{0}^{1}\pars{1 - y}^{a}y^{\pars{b - a}/2 - 1}\,\dd y \\[3mm]&\mbox{where}\qquad a > -1\,,\quad{b - a \over 2} - 1 > - 1\,,\quad b - a > 0\qquad\imp\qquad \begin{array}{|c|}\hline\ b > a > -1\ \\ \hline\end{array}\qquad\pars{3} \end{align}$$ $\pars{3}$ garantiza la convergencia de las integrales y la desaparición del segundo término en $\pars{2}$ : $$\begin{align}&\color{#c00000}{\int_{0}^{\pi/2} \cos^{a}\pars{\theta}\cos\pars{b\theta}\,\dd\theta} ={1 \over 2^{a + 1}}\,\sin\pars{\bracks{b - a}\pi \over 2} {\rm B}\pars{a + 1,{b - a \over 2}} \end{align}$$ $\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$ es el Función Beta y $\ds{\Gamma\pars{z}}$ es el Función gamma . $$\begin{align}&\color{#c00000}{\int_{0}^{\pi/2} \cos^{a}\pars{\theta}\cos\pars{b\theta}\,\dd\theta} ={1 \over 2^{a + 1}}\,\sin\pars{\bracks{b - a}\pi \over 2} {\Gamma\pars{a + 1}\Gamma\pars{\bracks{b - a}/2} \over \Gamma\pars{a/2 + b/2 + 1}} \end{align}$$ Con Fórmula de reflexión de Euler : $$\begin{align}&\color{#c00000}{\int_{0}^{\pi/2} \cos^{a}\pars{\theta}\cos\pars{b\theta}\,\dd\theta} ={1 \over 2^{a + 1}}\,\sin\pars{\bracks{b - a}\pi \over 2} {\Gamma\pars{a + 1} \over \Gamma\pars{\bracks{a + b}/2 + 1}}\times \\[3mm]&{\pi \over \sin\pars{\pi\braces{1 - \bracks{b - a}/2}} \Gamma\pars{1 - \bracks{b - a}/2}} \end{align}$$

$$\begin{align}&\color{#00f}{\large\int_{0}^{\pi/2} \cos^{a}\pars{\theta}\cos\pars{b\theta}\,\dd\theta} \\[3mm]&=\color{#00f}{\large{\pi \over 2^{a + 1}}\, {\Gamma\pars{a + 1}\over \Gamma\pars{1 + \bracks{a + b}/2}\Gamma\pars{1 + \bracks{a - b}/2}}}\,,\qquad \verts{b} > a > -1 \end{align}$$

$\ds{\Huge\tt ADDENDUM:}$

Con $\ds{0 < \mu < 1}$ : $$\begin{align}&\color{#66f}{\large\int_{0}^{\pi/2} \cos^{a}\pars{\theta}\cos\pars{a\theta}\,\dd\theta} =\Re\int_{0}^{\pi/2} \cos^{a}\pars{\theta}\expo{-\ic a\theta}\,\dd\theta \\[3mm]&=\lim_{\mu \to 1^{-}}\Re\int_{0}^{\pi/2} \pars{\expo{\ic\theta} + \mu\expo{-\ic\theta} \over 2}^{a} \expo{-\ic a\theta}\,\dd\theta ={1 \over 2^{a}}\lim_{\mu \to 1^{-}}\Re\int_{0}^{\pi/2} \pars{1 + \mu\expo{-2\ic\theta}}^{a}\,\dd\theta \\[3mm]&={1 \over 2^{a}}\lim_{\mu \to 1^{-}}\Re\sum_{n = 0}^{\infty} {a \choose n}\mu^{n}\int_{0}^{\pi/2}\expo{-2\ic n\theta}\,\dd\theta ={1 \over 2^{a}}\lim_{\mu \to 1^{-}}\sum_{n = 0}^{\infty} {a \choose n}\mu^{n}\ \overbrace{\int_{0}^{\pi/2}\cos\pars{2n\theta}\,\dd\theta} ^{\ds{=\ {\pi \over 2}\,\delta_{n0}}} \\[3mm]&={\pi \over 2^{a + 1}}\,\lim_{\mu \to 1^{-}}{a \choose 0}\mu^{0} =\color{#66f}{\Large{\pi \over 2^{a + 1}}} \end{align}$$

Answer 3

Mi suposición sería utilizar la identidad de Euler, y

$$\exp( i x \theta ) = \cos (x \theta) + i \sin (x\theta )$$

Esto debería reducir la integral a un trato de tipo exponencial.