Demostrar que $\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}$

Question

Yo estaba tratando de resolver la integral de la $$\int_{0}^{\infty}\frac{x^4}{1+x^6}dx$$ utilizando la serie. Ahora estoy atascado en la serie a continuación.

Cómo probar que

$$\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}?$$

Answer 1

Denotar $$ S = \sum_{j=0}^{\infty}(-1)^j \left( \frac{1}{6j+1} + \dfrac{1}{6j+5}\right). $$ Y denotan $$ T= \sum_{j=0}^{\infty} (-1)^j \dfrac{1}{2j+1}. $$ De acuerdo a la fórmula de Leibniz para $\pi$, $\;T=\dfrac{\pi}{4}$. Entonces $$S - \dfrac{1}{3}T = \\ S - \dfrac{1}{3}\sum_{j=0}^{\infty} (-1)^j \dfrac{1}{2j+1} = \\ S-\sum_{j=0}^{\infty}(-1)^{j}\left(\dfrac{1}{6j+3}\right) = \\ \sum_{j=0}^{\infty}(-1)^j \left( \frac{1}{6j+1} - \frac{1}{6j+3} + \dfrac{1}{6j+5}\right) = \\ \sum_{k=0}^{\infty} (-1)^k \dfrac{1}{2k+1} = \\ T. $$

Por lo tanto, $$S-\dfrac{1}{3}T = T,$$ $$S=\dfrac{4}{3}T,$$ $$S=\dfrac{4}{3}\cdot \dfrac{\pi}{4}=\dfrac{\pi}{3}.$$

Answer 2

$$ \begin{align} \sum_{j=0}^\infty(-1)^j\left(\frac1{5+6j}+\frac1{1+6j}\right) &=\sum_{j=0}^\infty(-1)^j\left(\frac1{-1-6(-j-1)}+\frac1{1+6j}\right)\tag1\\ &=\sum_{j=0}^\infty(-1)^{-j-1}\frac1{1+6(-j-1)}+\sum_{j=0}^\infty(-1)^j\frac1{1+6j}\tag2\\ &=\sum_{j=-\infty}^{-1}(-1)^j\frac1{1+6j}+\sum_{j=0}^\infty(-1)^j\frac1{1+6j}\tag3\\ &=\frac16\sum_{j=-\infty}^\infty(-1)^j\frac1{\frac16+j}\tag4\\[6pt] &=\frac\pi6\csc\left(\frac\pi6\right)\tag5\\[12pt] &=\frac\pi3\tag6 \end{align} $$ Explicación:
$(1)$: $5+6j=-1-6(-j-1)$
$(2)$: dividir la suma de dos
$(3)$: sustituto $j\mapsto-1-j$
$(4)$: combinar las sumas
$(5)$: el uso de $(3)$ a partir de esta respuesta
$(6)$: evaluar

Answer 3

$\begin{align}J&=\int_0^{\infty}\frac{x^4}{1+x^6}\,dx\\ &=\int_0^{\infty}\frac{2x^2-1}{3(x^4-x^2+1)}\,dx+\frac{1}{3}\int_0^{\infty}\frac{1}{1+x^2}\,dx\\ &=\int_0^{\infty}\frac{2x^2-1}{3(x^4-x^2+1)}\,dx+\frac{1}{3}\Big[\arctan x\Big]_0^{\infty}\\ &=\frac{1}{3}\int_0^{\infty}\frac{2x^2-1}{x^4-x^2+1}\,dx+\frac{1}{6}\pi \end{align}$

Realizar el cambio de variable $y=\dfrac{1}{x}$,

$\begin{align}K&=\int_0^{\infty}\frac{2x^2-1}{x^4-x^2+1}\,dx\\ &=\int_0^{\infty}\frac{2-x^2}{x^4-x^2+1}\,dx \end{align}$

Por lo tanto,

$\begin{align}2K&=\int_0^{\infty}\frac{(2x^2-1)+(2-x^2)}{x^4-x^2+1}\,dx\\ &=\int_0^{\infty}\frac{1+x^2}{x^4-x^2+1}\,dx\\ &=\int_{0}^{+\infty}\frac{\frac{1}{x^2}+1}{x^2-1+\frac{1}{x^2}}\,dx\\ &=\int_{0}^{+\infty}\frac{\frac{1}{x^2}+1}{\left(x-\frac{1}{x}\right)^2+1}\,dx \end{align}$

Realizar el cambio de variable $y=x-\dfrac{1}{x}$,

$\begin{align}2K&=\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\,dx\\ &=\Big[\arctan x\Big]_{-\infty}^{+\infty}\\ &=\pi \end{align}$

Por lo tanto,

$\begin{align}K&=\frac{1}{2}\pi\end{align}$

Por lo tanto,

$\begin{align}J&=\frac{1}{3}K+\frac{1}{6}\pi\\ &=\frac{1}{3}\times \frac{1}{2}\pi++\frac{1}{6}\pi\\ &=\boxed{\frac{1}{3}\pi} \end{align}$