El cálculo de $ \int ^{ \pi /2}_{0} \cos ^{n}(x) \cos (nx)\,dx$ donde $n \in \mathbb {N}$

Question

Calcular la integral definitiva

$$ \int ^{ \pi /2}_{0} \cos ^{n}(x) \cos (nx)\,dx $$

donde $n \in \mathbb {N}$ .

Mi intento:

Usando $ \cos (x) = \frac {e^{ix}+e^{-ix}}{2}$ tenemos

$$ \begin {align} \int ^{ \pi /2}_{0} \cos ^{n}(x) \cos (nx)\,dx&= \int_ {0}^{ \pi /2} \left ( \frac {e^{ix}+e^{-ix}}{2} \right )^n \left ( \frac {e^{inx}+e^{-inx}}{2} \right )\,dx \\ &= \frac {1}{2^n} \mathrm {Re} \left\ { \int_ {0}^{ \pi /2} \left (e^{ix}+e^{-ix} \right )^n \cdot e^{inx}\,dx \right\ } \\ &= \frac {1}{2^n} \mathrm {Re} \left\ { \int_ {0}^{ \pi /2} \left (e^{2ix}+1 \right )^n\,dx \right\ } \end {align} $$

Dejando $z=e^{4ix}$ nos da

$$ \begin {align} e^{4ix}dx &= \frac {1}{4i}dz \\ dx &= \frac {dz}{4iz} \end {align} $$

Así que la integral se convierte

$$ \frac {1}{4 \cdot 2^n} \mathrm {Re} \left\ { \int_ {C} \left ( \sqrt {z}+1 \right ) \cdot \frac {dz}{iz} \right\ }$$

¿Cómo puedo completar la solución desde aquí?

Answer 1

Tu comienzo no es malo, pero puedes conseguirlo más simple usando la uniformidad de $ \cos $ ,

$$ \int_0 ^{ \pi /2} \cos ^n x \cos (nx)\,dx = \frac12 \int_ {- \pi /2}^{ \pi /2} \cos ^n x \cos (nx)\,dx.$$

Dado que el seno es una función impar, podemos reemplazar $ \cos (nx)$ con $e^{inx}$ :

$$ \begin {align} \int_0 ^{ \pi /2} \cos ^n x \cos (nx)\,dx &= \frac12\int_ {- \pi /2}^{ \pi /2} \cos ^n x \, e^{inx}\,dx \\ &= \frac {1}{2^{n+1}} \int_ {- \pi /2}^{ \pi /2} \left (e^{ix}+e^{-ix} \right )^n e^{inx}\,dx \\ &= \frac {1}{2^{n+1}} \int_ {- \pi /2}^{ \pi /2} \left (e^{2ix}+1 \right )^n\,dx. \end {align}$$

A continuación sustituimos $ \varphi = 2x$ y conseguir

$$ \int_0 ^{ \pi /2} \cos ^n x \cos (nx)\,dx = \frac {1}{2^{n+2}} \int_ {- \pi }^ \pi \left (e^{i \varphi }+1 \right )^n\,d \varphi. $$

Ahora el ajuste $z = e^{i \varphi }$ nos da un bonito contorno integral sobre el círculo de la unidad,

$$ \begin {align} \int_0 ^{ \pi /2} \cos ^n x \cos (nx)\,dx &= \frac {1}{2^{n+2}} \int_ { \lvert z \rvert = 1} (z+1)^n\, \frac {dz}{iz} \\ &= \frac { \pi }{2^{n+1}} \end {align}$$

por la fórmula integral de Cauchy.

Answer 2

Pista:

Ponga $$I_n= \int_ {0}^{ \frac { \pi }{2}} \cos ^nx \cos (nx)dx= \int_ {0}^{ \frac { \pi }{2}} \cos ^{n-1}x \cos [(n-1)x]dx- \int_ {0}^{ \frac { \pi }{2}} \cos ^{n-1}x \sin (nx) \sin xdx=I_{n-1}+ \int_ {0}^{ \frac { \pi }{2}} \sin (nx)d( \frac { \cos ^nx}{n})=I_{n-1}-I_{n} $$

$$ \to I_n= \frac {1}{2}I_{n-1}= \cdots = \frac {1}{2^{n}}I_0= \frac { \pi }{2^{n+1}}$$

Answer 3

$ \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 { \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 { \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 { \verts }[1]{ \left\vert\ , #1 \, \right\vert }$ $ \ds { \int_ {0}^{ \pi /2} \cos ^{n} \pars {x} \cos\pars {nx}\, \dd x:\ { \large ?}. \quad n \in { \mathbb N}}$ .

Con $ \ds {0 < \epsilon < 1}$ $ \ds { \pars {~ \mbox {we'll take at the end the limit}\ \epsilon \to 0^{+}~}}$ : \begin {alineado}& \color {#66f}{ \large\int_ {0}^{ \pi /2} \cos ^{n} \pars {x} \cos\pars {nx} \, \dd x}= \Re \int_ { \verts {z}\ =\ 1 \atop { \vphantom { \Huge A}0\ <\ { \rm Arg} \pars {z}\ <\ \pi /2}} \pars {z^{2} + 1 \over \dd z \over \ic z} \\ [3mm]&={1 \over 2^{n}}\, \Im \int_ { \verts {z}\ =\ 1 \atop { \vphantom { \Huge A}0\ <\ { \rm Arg} \pars {z}\ <\ \pi /2}} { \pars {z^{2} + 1}^{n} \over z}\, \dd z =-\,{1 \over 2^{n}}\, \Im\int_ {1}^{ \epsilon } { \pars {-y^{2} + 1}^{n} \over \ic y}\, \ic\ , \dd y \\ [3mm]& \phantom {=} \left.\mbox {}-{1 \over 2^{n}}\, \Im\int_ { \pi /2}^{0} { \pars {z^{2} + 1}^{n} \over z}\, \dd z\, \right\vert_ {\,z\ =\ \epsilon\expo { \ic\theta }} -{1 \over 2^{n}}\, \Im\int_ { \epsilon }^{1} { \pars {x^{2} + 1}^{n} \over x}\, \dd x \\ [3mm]&=-\,{1 \over 2^{n}} \Im\int_ { \pi /2}^{0} \ic\ , \dd\theta = \color {#66f}{ \Large { \pi \over 2^{n + 1}}} \end {alinear}