How i could prove this:$F(a)=\int_{0}^{\pi/2} \cos^{a}\! \theta \cos(a\theta){\mathrm{d}}\theta$= $\frac{\pi}{2^{a+1}}, \quad$

Question

for $ a>-1$

prove that:

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

I would be interest for any replies or any comments

Answer 1

I used this formula that Evaluated by Augustin Louis Cauchy:

for $a>-1$ and $0 \leq b<a+2$:

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

it's seems to me for a=b as a special case i get :

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

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}

When $\ds{\epsilon \to 0^{+}}$ I find $\ds{\pars{~\mbox{from}\ \pars{1}\ \mbox{and}\ \pars{2}~}}$: \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} $\pars{3}$ guarantees the integrals convergence and the vanishing out of the second term in $\pars{2}$: \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} $\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$ is the Beta Function and $\ds{\Gamma\pars{z}}$ is the Gamma Function. \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} With Euler Reflection Formula: \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}

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

With $\ds{0 < \mu < 1}$: \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

My guess would be to use Euler's identity, and

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

This should reduce the integral to an exponential type deal.

Answer 4

Credits to @Felix Marin $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \cos ^a \theta \cos (a \theta) d \theta & =\Re\int_0^\frac\pi2 \cos ^a \theta e^{a \theta i} d \theta \\ & =\Re\int_0^ \frac\pi2\left(\frac{e^{\theta i}+e^{-\theta i}}{2}\right)^a e^{a \theta i} d \theta \\ & =\frac{1}{2^a} \Re\int_0^{\frac{\pi}{2}}\left(e^{2 \theta i}+1\right)^a d \theta \\ & =\frac{1}{2^a} \Re\sum_{n=0}^{\infty}\binom{a}{n} \int_0^{\frac{\pi}{2}} e^{2 n \theta i} d \theta \\ & =\frac{1}{2^a} \sum_{n=0}^{\infty}\binom{a}{n} \int_0^{\frac{\pi}{2}} \cos (2 n \theta) d \theta \\ & =\frac{1}{2^a}\left(\binom{a}{0}\int_0^{\frac{\pi}{2}} d \theta+\sum_{n=1}^{\infty}\binom{a}{n}\left[\frac{\sin (2 n \theta)}{2 n}\right]_0^{\frac{\pi}{2}}\right) \\ & =\frac{\pi}{2^{a+1}} \end{aligned} $$