Linealizar Potencias Enteras De Coseno

La expresión general 2^{n}\cos ^{n}\left( x\right) que es una potencia de la función trigonométrica coseno, es reducible a una suma de cosenos lineales de acuerdo a la siguiente fórmula.

Proposición:

2^{n}\cos ^{n}\left( x\right) =\left\{ \begin{array}{cc} \sum\limits_{i=0}^{\left\lceil \frac{n-1}{2}\right\rceil }\binom{n}{i}2\cos\left( \left( n-2i\right) x\right) & n\text{ impar} \\ \sum\limits_{i=0}^{\left\lceil \frac{n-1}{2}\right\rceil -1}\binom{n}{i} 2\cos \left( \left( n-2i\right) x\right) +2\binom{n-1}{\left\lceil \frac{n-1}{2}\right\rceil -1} & n\text{ par} \end{array} \right.

La demostración se hará por inducción para n par e impar, demostrando pares e impares por separado.


DEMOSTRACIÓN

Con n impar

Para n=1 tenemos:

2^{1}\cos ^{1}\left( x\right) =\sum\limits_{i=0}^{\left\lceil \frac{1-1}{2}\right\rceil }\binom{1}{i}2\cos \left( \left( 1-2i\right) x\right) =2\cos \left( x\right)

Suponemos para n=2p-1

2^{2p-1}\cos ^{2p-1}\left( x\right) =\sum\limits_{i=0}^{p-1}\binom{2p-1}{i}2\cos \left( \left( 2p-1-2i\right) x\right)

Para demostrar el siguiente impar multiplicamos por 2^{2}\cos ^{2}\left( x\right)

\left( 2^{2}\cos ^{2}\left( x\right) \right) \left( 2^{2p-1}\cos ^{2p-1}\left( x\right) \right) =\left( 2^{2}\cos ^{2}\left( x\right) \right) \sum\limits_{i=0}^{p-1}\binom{2p-1}{i}2\cos \left( \left( 2p-1-2i\right) x\right)

Introduciendo a la sumatoria y viendo que 2^{2}\cos ^{2}\left( x\right) =2\left( \cos 2x+1\right)

=\sum\limits_{i=0}^{p-1}\binom{2p-1}{i}2\left( \cos \left( \left( 2p-1-2i\right) x\right) \left( 2\left( \cos 2x+1\right) \right) \right)

Multiplicando

=\sum\limits_{i=0}^{p-1}\binom{2p-1}{i}2\left( 2\cos \left( \left( 2p-1-2i\right) x\right) \left( \cos 2x\right) +2\cos \left( \left( 2p-1-2i\right) x\right) \right)

Utilizando la identidad trigonométrica 2\cos x\cos y=\cos \left( x+y\right) +\cos \left( x-y\right)

=\sum\limits_{i=0}^{p-1}\binom{2p-1}{i}2\left( \cos \left( \left( 2p-1-2i+2\right) x\right) +\cos \left( \left( 2p-1-2i-2\right) x\right) +2\cos \left( \left( 2p-1-2i\right) x\right) \right)

Separando sumatorias

=\sum\limits_{i=0}^{p-1}\binom{2p-1}{i}2\cos \left( \left( 2p-1-2i+2\right) x\right) +\sum\limits_{i=0}^{p-1}\binom{2p-1}{i}2\cos \left( \left(2p-1-2i-2\right) x\right) +\sum\limits_{i=0}^{p-1}\binom{2p-1}{i}2^{2}\cos\left( \left( 2p-1-2i\right) x\right)

Para poder sumar correctamente vamos a separar los dos primeros terminos de la primera sumatoria, los dos últimos terminos de la segunda sumatoria, el primero y el último termino de la tercera sumatoria. De este modo:

=\sum\limits_{i=2}^{p-1}\binom{2p-1}{i}2\cos \left( \left( 2p-1-2i+2\right) x\right) +\sum\limits_{i=0}^{p-3}\binom{2p-1}{i}2\cos \left( \left( 2p-1-2i-2\right) x\right) +\sum\limits_{i=1}^{p-2}\binom{2p-1}{i}2^{2}\cos \left( \left( 2p-1-2i\right) x\right)

+\binom{2p-1}{0}2\cos \left( \left( 2p+1\right) x\right) +\binom{2p-1}{1}2\cos \left( \left( 2p-1\right) x\right) +\binom{2p-1}{p-2}2\cos \left( \left( 1\right) x\right) +\binom{2p-1}{p-1}2\cos \left( \left( -1\right) x\right)

+\binom{2p-1}{0}2^{2}\cos \left( \left( 2p-1\right) x\right) +\binom{2p-1}{p-1}2^{2}\cos \left( \left( 1\right) x\right)

Ahora simplemente renombramos y en los terminos separados sumamos recordando que \cos x=\cos \left( -x\right)

=\sum\limits_{j=0}^{p-3}\binom{2p-1}{j+2}2\cos \left( \left( 2p-3-2j\right) x\right) +\sum\limits_{j=0}^{p-3}\binom{2p-1}{j}2\cos \left( \left( 2p-3-2j\right) x\right) +\sum\limits_{j=0}^{p-3}\binom{2p-1}{j+1}2^{2}\cos \left( \left( 2p-3-2j\right) x\right)

+\binom{2p-1}{0}2\cos \left( \left( 2p+1\right) x\right) +\left( \binom{2p-1}{1}+2\binom{2p-1}{0}\right) 2\cos \left( \left( 2p-1\right) x\right) +\left( \binom{2p-1}{p-2}+\binom{2p-1}{p-1}+2\binom{2p-1}{p-1}\right) 2\cos \left( \left( 1\right) x\right)

Factorizando

=\sum\limits_{j=0}^{p-3}\left( \binom{2p-1}{j+2}+\binom{2p-1}{j}+2\binom{2p-1}{j+1}\right) 2\cos \left( \left( 2p-3-2j\right) x\right)

+\binom{2p-1}{0}2\cos \left( \left( 2p+1\right) x\right) +\left( 2p-1+2\left( 1\right) \right) 2\cos \left( \left( 2p-1\right) x\right) +\left( \frac{\left( 2p-1\right) !}{\left( p-2\right) !\left( p+1\right) !}+3\frac{\left( 2p-1\right) !}{\left( p-1\right) !p!}\right) 2\cos \left( \left( 1\right) x\right)

Reexpresando convenientemente

=\sum\limits_{j=0}^{p-3}\left( \frac{\left( 2p-1\right) !}{\left( j+2\right) !\left( 2p-3-j\right) !}+\frac{\left( 2p-1\right) !}{j!\left( 2p-1-j\right) !}+2\frac{\left( 2p-1\right) !}{\left( j+1\right) !\left( 2p-2-j\right) !}\right) 2\cos \left( \left( 2p-3-2j\right) x\right)

+\binom{2p-1}{0}2\cos \left( \left( 2p+1\right) x\right) +\left( 2p+1\right) 2\cos \left( \left( 2p-1\right) x\right) +\left( 2p-1\right) !\left( \frac{1}{\left( p-2\right) !\left( p+1\right) !}+3\frac{1}{\left( p-1\right) !p!}\right) 2\cos \left( \left( 1\right) x\right)

=\sum\limits_{j=0}^{p-3}\frac{\left( 2p-1\right) !}{j!\left( 2p-1-j\right) !}\left( \frac{\left( 2p-1-j\right) \left( 2p-2-j\right) }{\left( j+2\right) \left( j+1\right) }+1+2\frac{\left( 2p-1-j\right) }{\left( j+1\right) }\right) 2\cos \left( \left( 2p-3-2j\right) x\right)

+\binom{2p-1}{0}2\cos \left( \left( 2p+1\right) x\right) +\left( 2p+1\right) 2\cos \left( \left( 2p-1\right) x\right) +\frac{\left( 2p-1\right) !}{p!\left( p-1\right) !}\left( \frac{\left( p-1\right) }{\left( p+1\right) }+3\right) 2\cos \left( \left( 1\right) x\right)

Sumando

=\sum\limits_{j=0}^{p-3}\frac{\left( 2p-1\right) !}{j!\left( 2p-1-j\right) !}\left( \frac{\left( 2p\right) \left( 2p+1\right) }{\left( j+2\right) \left( j+1\right) }\right) 2\cos \left( \left( 2p-3-2j\right) x\right)

+\binom{2p-1}{0}2\cos \left( \left( 2p+1\right) x\right) +\left( 2p+1\right) 2\cos \left( \left( 2p-1\right) x\right) +\frac{\left( 2p-1\right) !}{p!\left( p-1\right) !}\left( \frac{2\left( 2p+1\right) }{\left( p+1\right) }\right) 2\cos \left( \left( 1\right) x\right)

Reexpresando

=\sum\limits_{j=0}^{p-3}\left( \frac{\left( 2p+1\right) !}{\left( j+2\right) !\left( 2p-1-j\right) !}\right) 2\cos \left( \left( 2p-3-2j\right) x\right)

+\binom{2p-1}{0}2\cos \left( \left( 2p+1\right) x\right) +\left( 2p+1\right) 2\cos \left( \left( 2p-1\right) x\right) +\frac{\left( 2p+1\right) !}{\left( p+1\right) !\left( p-1\right) !}\left( \frac{2}{\left( 2p\right) }\right) 2\cos \left( \left( 1\right) x\right)

Reexpresando convenientemente

=\sum\limits_{j=0}^{p-3}\binom{2p+1}{j+2}2\cos \left( \left( 2p-3-2j\right) x\right) +\binom{2p+1}{0}2\cos \left( \left( 2p+1\right) x\right) +\binom{2p+1}{1}2\cos \left( \left( 2p-1\right) x\right) +\binom{2p+1}{p}2\cos \left( \left( 1\right) x\right)

Reescribiendo el índice de la sumatoria

=\sum\limits_{i=2}^{p-1}\binom{2p+1}{i}2\cos \left( \left( 2p+1-2i\right) x\right) +\binom{2p+1}{0}2\cos \left( \left( 2p+1\right) x\right) +\binom{2p+1}{1}2\cos \left( \left( 2p+1-2\left( 1\right) \right) x\right) +\binom{2p+1}{p}2\cos \left( \left( 2p+1-2\left( p\right) \right) x\right)

Incorporando los terminos separados

=\sum\limits_{i=0}^{p}\binom{2p+1}{i}2\cos \left( \left( 2p+1-2i\right) x\right)

=2^{2p+1}\cos ^{2p+1}\left( x\right)

Demostrado para n impar.


Con n par

Para n=2 tenemos:

2^{2}\cos ^{2}\left( x\right) =\sum\limits_{i=0}^{\left\lceil \frac{2-1}{2}\right\rceil -1}\binom{2}{i}2\cos \left( \left( 2-2i\right) x\right) +2\binom{2-1}{\left\lceil \frac{2-1}{2}\right\rceil -1}=2\cos \left( 2x\right) +2

Suponemos para n=2p

2^{2p}\cos ^{2p}\left( x\right) =\sum\limits_{i=0}^{p-1}\binom{2p}{i}2\cos \left( \left( 2p-2i\right) x\right) +2\binom{2p-1}{p-1}

Para demostrar el siguiente par multiplicamos por 2^{2}\cos ^{2}\left( x\right)

\left( 2^{2}\cos ^{2}\left( x\right) \right) \left( 2^{2p}\cos ^{2p}\left( x\right) \right) =\left( 2^{2}\cos ^{2}\left( x\right) \right) \left( \sum\limits_{i=0}^{p-1}\binom{2p}{i}2\cos \left( \left( 2p-2i\right) x\right) +2\binom{2p-1}{p-1}\right)

Introduciendo a la sumatoria y viendo que 2^{2}\cos ^{2}\left( x\right) =2\left( \cos 2x+1\right)

=\sum\limits_{i=0}^{p-1}\binom{2p}{i}2\left( \cos \left( \left( 2p-2i\right) x\right) \left( 2\left( \cos 2x+1\right) \right) \right) +2 \binom{2p-1}{p-1}\left( 2\cos 2x+2\right)

Multiplicando

=\sum\limits_{i=0}^{p-1}\binom{2p}{i}2\left( 2\cos \left( \left( 2p-2i\right) x\right) \left( \cos 2x\right) +2\cos \left( \left( 2p-2i\right) x\right) \right) +2\binom{2p-1}{p-1}\left( 2\cos 2x+2\right)

Utilizando la identidad trigonométrica 2\cos x\cos y=\cos \left( x+y\right) +\cos \left( x-y\right)

=\sum\limits_{i=0}^{p-1}\binom{2p}{i}2\left( \cos \left( \left( 2p-2i+2\right) x\right) +\cos \left( \left( 2p-2i-2\right) x\right) +2\cos \left( \left( 2p-2i\right) x\right) \right) +2\binom{2p-1}{p-1}\left( 2\cos 2x+2\right)

Separando sumatorias

=\sum\limits_{i=0}^{p-1}\binom{2p}{i}2\cos \left( \left( 2p-2i+2\right) x\right) +\sum\limits_{i=0}^{p-1}\binom{2p}{i}2\cos \left( \left( 2p-2i-2\right) x\right) +\sum\limits_{i=0}^{p-1}\binom{2p}{i}2^{2}\cos \left( \left( 2p-2i\right) x\right) +2\binom{2p-1}{p-1}\left( 2\cos 2x+2\right)

Para poder sumar correctamente vamos a separar los dos primeros terminos de la primera sumatoria, los dos últimos terminos de la segunda sumatoria, el primero y el último termino de la tercera sumatoria. De este modo:

=\sum\limits_{i=2}^{p-1}\binom{2p}{i}2\cos \left( \left( 2p-2i+2\right) x\right) +\sum\limits_{i=0}^{p-3}\binom{2p}{i}2\cos \left( \left( 2p-2i-2\right) x\right) +\sum\limits_{i=1}^{p-2}\binom{2p}{i}2^{2}\cos \left( \left( 2p-2i\right) x\right)

+\binom{2p}{0}2\cos \left( \left( 2p+2\right) x\right) +\binom{2p}{1}2\cos \left( 2px\right) +\binom{2p}{p-2}2\cos \left( 2x\right) +\binom{2p}{p-1}2\cos \left( 0x\right)

+\binom{2p}{0}2^{2}\cos \left( 2px\right) +\binom{2p}{p-1}2^{2}\cos \left( 2x\right) +2\binom{2p-1}{p-1}\left( 2\cos 2x+2\right)

Ahora simplemente renombramos y en los terminos separados sumamos recordando que \cos 0x=1

=\sum\limits_{j=0}^{p-3}\binom{2p}{j+2}2\cos \left( \left( 2p-2-2j\right) x\right) +\sum\limits_{j=0}^{p-3}\binom{2p}{j}2\cos \left( \left( 2p-2-2j\right) x\right) +\sum\limits_{j=0}^{p-3}\binom{2p}{j+1}2^{2}\cos \left( \left( 2p-2-2j\right) x\right)

+\binom{2p}{0}2\cos \left( \left( 2p+2\right) x\right) +\left( \binom{2p}{1}+2\binom{2p}{0}\right) 2\cos \left( 2px\right) +\left( \binom{2p}{p-2}+2\binom{2p}{p-1}+2\binom{2p-1}{p-1}\right) 2\cos \left( 2x\right) +2\binom{2p}{p-1}+2^{2}\binom{2p-1}{p-1}

Factorizando

=\sum\limits_{j=0}^{p-3}\left( \binom{2p}{j+2}+\binom{2p}{j}+2\binom{2p}{j+1}\right) 2\cos \left( \left( 2p-2-2j\right) x\right)

+\binom{2p}{0}2\cos \left( \left( 2p+2\right) x\right) +\left( 2p+2\left( 1\right) \right) 2\cos \left( 2px\right) +\left( \frac{\left( 2p\right) !}{\left( p-2\right) !\left( p+2\right) !}+2\frac{\left( 2p\right) !}{\left( p-1\right) !\left( p+1\right) !}+2\frac{\left( 2p-1\right) !}{\left( p-1\right) !p!}\right) 2\cos \left( 2x\right) +2\left( \frac{\left( 2p\right) !}{\left( p-1\right) !\left( p+1\right) !}+2\frac{\left( 2p-1\right) !}{\left( p-1\right) !p!}\right)

Reexpresando convenientemente

=\sum\limits_{j=0}^{p-3}\left( \frac{\left( 2p\right) !}{\left( j+2\right) !\left( 2p-2-j\right) !}+\frac{\left( 2p\right) !}{j!\left( 2p-j\right) !}+2\frac{\left( 2p\right) !}{\left( j+1\right) !\left( 2p-1-j\right) !}\right) 2\cos \left( \left( 2p-2-2j\right) x\right)

+\binom{2p}{0}2\cos \left( \left( 2p+2\right) x\right) +\left( 2p+2\right) 2\cos \left( 2px\right) +\frac{\left( 2p\right) !}{\left( p-1\right) !\left( p+1\right) !}\left( \frac{\left( p-1\right) }{\left( p+2\right) }+2+2\frac{\left( p+1\right) }{\left( 2p\right) }\right) 2\cos \left( 2x\right) +2\frac{\left( 2p\right) !}{\left( p-1\right) !\left( p+1\right) !}\left( 1+2\frac{\left( p+1\right) }{\left( 2p\right) }\right)

=\sum\limits_{j=0}^{p-3}\frac{\left( 2p\right) !}{j!\left( 2p-j\right) !}\left( \frac{\left( 2p-j\right) \left( 2p-1-j\right) }{\left( j+2\right) \left( j+1\right) }+1+2\frac{\left( 2p-j\right) }{\left( j+1\right) }\right) 2\cos \left( \left( 2p-2-2j\right) x\right)

+\binom{2p}{0}2\cos \left( \left( 2p+2\right) x\right) +\left( 2p+2\right) 2\cos \left( 2px\right) +\frac{\left( 2p\right) !}{\left( p-1\right) !\left( p+1\right) !}\left( \frac{\left( 2p+1\right) \left( 2p+2\right) }{p\left( p+2\right) }\right) 2\cos \left( 2x\right) +2\frac{\left( 2p\right) !}{\left( p-1\right) !\left( p+1\right) !}\left( \frac{2p+1}{p}\right)

Sumando

=\sum\limits_{j=0}^{p-3}\frac{\left( 2p\right) !}{j!\left( 2p-j\right) !}\left( \frac{\left( 2p+1\right) \left( 2p+2\right) }{\left( j+2\right) \left( j+1\right) }\right) 2\cos \left( \left( 2p-2-2j\right) x\right)

+\binom{2p}{0}2\cos \left( \left( 2p+2\right) x\right) +\left( 2p+2\right) 2\cos \left( 2px\right) +\left( \frac{\left( 2p+2\right) !}{p!\left( p+2\right) !}\right) 2\cos \left( 2x\right) +2\left( \frac{\left( 2p+1\right) !}{p!\left( p+1\right) !}\right)

Reexpresando

=\sum\limits_{j=0}^{p-3}\left( \frac{\left( 2p+2\right) !}{\left( j+2\right) !\left( 2p-j\right) !}\right) 2\cos \left( \left( 2p-2-2j\right) x\right)

+\binom{2p}{0}2\cos \left( \left( 2p+2\right) x\right) +\left( 2p+2\right) 2\cos \left( 2px\right) +\left( \frac{\left( 2p+2\right) !}{p!\left( p+2\right) !}\right) 2\cos \left( 2x\right) +2\left( \frac{\left( 2p+1\right) !}{p!\left( p+1\right) !}\right)

Reexpresando convenientemente

=\sum\limits_{j=0}^{p-3}\binom{2p+2}{j+2}2\cos \left( \left( 2p-2-2j\right) x\right) +\binom{2p+2}{0}2\cos \left( \left( 2p+2\right) x\right) +\binom{2p+2}{1}2\cos \left( 2px\right) +\binom{2p+2}{p}2\cos \left( 2x\right) +2\binom{2p+1}{p}

Reescribiendo el índice de la sumatoria

=\sum\limits_{i=2}^{p-1}\binom{2p+2}{i}2\cos \left( \left( 2p+2-2i\right) x\right) +\binom{2p+2}{0}2\cos \left( \left( 2p+2\right) x\right) +\binom{2p+2}{1}2\cos \left( \left( 2p+2-2\left( 1\right) \right) x\right)

+\binom{2p+2}{p}2\cos \left( \left( 2p+2-2\left( p\right) \right) x\right) +2\binom{2p+1}{\left( p+1\right) -1}

Incorporando los terminos separados

=\sum\limits_{i=0}^{p}\binom{2p+2}{i}2\cos \left( \left( 2p+2-2i\right) x\right) +2\binom{2p+1}{\left\lceil \frac{2p+1}{2}\right\rceil -1}

=2^{2p+2}\cos ^{2p+2}\left( x\right)

Demostrado para n par.


Post #11mycomplexsoul

Integrales Logarítmicas

Para n\geq 0 entero, se tiene

\int \left( \ln x\right) ^{n}dx=x\sum\limits_{i=0}^{n}\left( -1\right) ^{i}\frac{n!}{\left( n-i\right) !}\left( \ln x\right) ^{n-i}

Donde 0!=1. La demostración se hace por inducción.


DEMOSTRACIÓN

Para n=0. Muy simple, pero también se cumple.

\int \left( \ln x\right) ^{0}dx=x\sum\limits_{i=0}^{0}\left( -1\right) ^{i}\frac{0!}{\left( 0-i\right) !}\left( \ln x\right) ^{0-i}=x\left( 1\right) =x

Para n=1.

\int \left( \ln x\right) ^{1}dx=x\sum\limits_{i=0}^{1}\left( -1\right) ^{i}\frac{1!}{\left( 1-i\right) !}\left( \ln x\right) ^{1-i}=x\left( \ln x-1\right)

Suponemos para n\leq k.

\int \left( \ln x\right) ^{k}dx=x\sum\limits_{i=0}^{k}\left( -1\right) ^{i}\frac{k!}{\left( k-i\right) !}\left( \ln x\right) ^{k-i}

Probamos para n=k+1. Integrando por partes.

\int \left( \ln x\right) ^{k+1}dx=\int \left( \ln x\right) ^{k}\ln xdx

=x\left( \ln x-1\right) \left( \ln x\right) ^{k}-k\int \left( \left( \ln x\right) ^{k}-\left( \ln x\right) ^{k-1}\right) dx

=x\left( \ln x-1\right) \left( \ln x\right) ^{k}-k\int \left( \ln x\right) ^{k}dx+k\int \left( \ln x\right) ^{k-1}dx

Sustituyendo

\int \left( \ln x\right) ^{k+1}dx=x\left( \ln x-1\right) \left( \ln x\right) ^{k}-kx\sum\limits_{i=0}^{k}\left( -1\right) ^{i}\frac{k!}{\left( k-i\right) !}\left( \ln x\right) ^{k-i}
+kx\sum\limits_{i=0}^{k-1}\left( -1\right) ^{i}\frac{\left( k-1\right) !}{\left( k-1-i\right) !}\left( \ln x\right) ^{k-1-i}

Sacando el primer termino de la primera sumatoria y renombrando índices para sumar adecuadamente

\int \left( \ln x\right) ^{k+1}dx=x\left( \ln x-1\right) \left( \ln x\right) ^{k}-kx\left( \ln x\right) ^{k}
-kx\sum\limits_{j=0}^{k-1}\left( \left( -1\right) ^{j-1}\frac{k!}{\left( k-1-j\right) !}\left( \ln x\right) ^{k-1-j}-\left( -1\right) ^{j}\frac{\left( k-1\right) !}{\left( k-1-j\right) !}\left( \ln x\right) ^{k-1-j}\right)

Acomodando

\int \left( \ln x\right) ^{k+1}dx=x\left( \ln x-1\right) \left( \ln x\right) ^{k}-kx\left( \ln x\right) ^{k}
-x\sum\limits_{j=0}^{k-1}\left( -1\right) ^{j-1}k\left( \frac{k!}{\left( k-1-j\right) !}+\frac{\left( k-1\right) !}{\left( k-1-j\right) !}\right) \left( \ln x\right) ^{k-1-j}

Simplificando

\int \left( \ln x\right) ^{k+1}dx=x\left( \ln x\right) ^{k+1}-\left( k+1\right) x\left( \ln x\right) ^{k}
+x\sum\limits_{j=0}^{k-1}\left( -1\right) ^{j-2}\frac{\left( k+1\right) !}{\left( k-1-j\right) !}\left( \ln x\right) ^{k-1-j}

Renombrando índice, 2 posiciones para incluir los dos terminos aislados

\int \left( \ln x\right) ^{k+1}dx=x\left( \ln x\right) ^{k+1}-\left( k+1\right) x\left( \ln x\right) ^{k}
+x\sum\limits_{i=2}^{k+1}\left( -1\right) ^{i}\frac{\left( k+1\right) !}{\left( k+1-i\right) !}\left( \ln x\right) ^{k+1-i}

Introduciendo en la sumatoria con i=0 y i=1 respectivamente

\int \left( \ln x\right) ^{k+1}dx=x\sum\limits_{i=0}^{k+1}\left( -1\right) ^{i}\frac{\left( k+1\right) !}{\left( k+1-i\right) !}\left( \ln x\right) ^{k+1-i}

Demostrado \forall n\in \mathbb{N}.


Post #06mycomplexsoul

Integrales Exponenciales

La integral exponencial simple \int x^{n}e^{ax}dx con n\geq 0 y a
constante tiene por resultado:

\int x^{n}e^{ax}dx=e^{ax}\sum\limits_{i=0}^{n}\left( -1\right) ^{i}\frac{1}{a^{i+1}}\frac{d^{i}}{dx^{i}}\left( x^{n}\right)

Donde \frac{d^{0}}{dx^{0}}\left( x^{n}\right) =x^{n}.

Para demostrar este resultado usamos inducción sobre n.


DEMOSTRACIÓN

Para n=0.

\int x^{0}e^{ax}dx=e^{ax}\sum\limits_{i=0}^{0}\left( -1\right) ^{i}\frac{1}{a^{i+1}}\frac{d^{i}}{dx^{i}}\left( x^{0}\right) =e^{ax}\left( \frac{1}{a}\right)

Suponemos para n=k.

\int x^{k}e^{ax}dx=e^{ax}\sum\limits_{i=0}^{k}\left( -1\right) ^{i}\frac{1}{a^{i+1}}\frac{d^{i}}{dx^{i}}\left( x^{k}\right)

Probamos para n=k+1. Usando integración por partes.

\int x^{k+1}e^{ax}dx=\frac{1}{a}x^{k+1}e^{ax}-\frac{\left( k+1\right) }{a}\int x^{k}e^{ax}dx

Sustituyendo

\int x^{k+1}e^{ax}dx=\frac{1}{a}x^{k+1}e^{ax}-\frac{\left( k+1\right) }{a}e^{ax}\sum\limits_{i=0}^{k}\left( -1\right) ^{i}\frac{1}{a^{i+1}}\frac{d^{i}}{dx^{i}}\left( x^{k}\right)

Reescribiendo el índice de la sumatoria

\int x^{k+1}e^{ax}dx=e^{ax}\left( \frac{1}{a}x^{k+1}-\frac{\left( k+1\right) }{a}\sum\limits_{j=1}^{k+1}\left( -1\right) ^{j-1}\frac{1}{a^{j}}\frac{d^{j-1}}{dx^{j-1}}\left( x^{k}\right) \right)

Reescribiendo

\int x^{k+1}e^{ax}dx=e^{ax}\left( \frac{1}{a}x^{k+1}+\sum\limits_{j=1}^{k+1}\left( -1\right) ^{j}\frac{1}{a^{j+1}}\frac{d^{j-1}}{dx^{j-1}}\left( \left( k+1\right) x^{k}\right) \right)

\int x^{k+1}e^{ax}dx=e^{ax}\left( \frac{1}{a}x^{k+1}+\sum\limits_{j=1}^{k+1}\left( -1\right) ^{j}\frac{1}{a^{j+1}}\frac{d^{j-1}}{dx^{j-1}}\left( \frac{d}{dx}\left( x^{k+1}\right) \right) \right)

\int x^{k+1}e^{ax}dx=e^{ax}\left( \frac{1}{a}x^{k+1}+\sum\limits_{j=1}^{k+1}\left( -1\right) ^{j}\frac{1}{a^{j+1}}\frac{d^{j}}{dx^{j}}\left( x^{k+1}\right) \right)

Introduciendo el termino restante con j=0.

\int x^{k+1}e^{ax}dx=e^{ax}\sum\limits_{j=0}^{k+1}\left( -1\right) ^{j}\frac{1}{a^{j+1}}\frac{d^{j}}{dx^{j}}\left( x^{k+1}\right)

Demostrado \forall n\in \mathbb{N}.


Post #02mycomplexsoul