Hoy, computando$\int_{0}^{\pi/2}\sin^2(\sin^2 x)\,dx$, encontré una identidad interesante:$$\sum_{k\geq 0}\frac{(-1)^k(4k)!}{(2k)!^3 4^k}=\cos(1)\cdot\sum_{k\geq 0}\frac{(-1)^k}{k!^2 4^k}.$ $ ¿Cómo lo demostraría?
Respuestas
¿Demasiados anuncios?
Otro enfoque puede ser este. Tenemos $$\cos\left(x\right)J_{0}\left(x\right)=\sum_{k\geq0}\frac{\left(-1\right)^{k}}{\left(2k\right)!}x^{2k}\sum_{k\geq0}\frac{\left(-1\right)^{k}}{4^{k}k!^{2}}x^{2k} $$ and by Cauchy product we get $$=\sum_{k\geq0}c_{k}\left(x\right) $$ where $$c_{k}\left(x\right)=\left(-1\right)^{k}x^{2k}\sum_{j=0}^{k}\frac{4^{-j}}{j!^{2}\left(2k-2j\right)!}. $$ Now observe that $$\sum_{j=0}^{k}\frac{4^{-j}}{j!^{2}\left(2k-2j\right)!}=\frac{1}{\left(2k\right)!}\sum_{j=0}^{k}\frac{\left(2k\right)!}{j!\left(2k-2j\right)!}\frac{4^{-j}}{j!}=$$ $$=\frac{1}{\left(2k\right)!}\sum_{j=0}^{k}\frac{\left(2k\right)\left(2k-2\right)\cdots\left(2k-2j+2\right)\left(2k-1\right)\cdots\left(2k-2j+1\right)}{j!}\frac{4^{-j}}{j!}=$$ $$=\frac{1}{\left(2k\right)!}\sum_{j=0}^{\infty}\frac{k\left(k-1\right)\cdots\left(k-j+1\right)\left(k-1/2\right)\cdots\left(k-j+1/2\right)}{j!}\frac{1}{j!}= $$ $$=\frac{1}{\left(2k\right)!}\sum_{j=0}^{\infty}\frac{\left(-k\right)_{j}\left(-k+1/2\right)_{j}}{j!}\frac{1}{j!}=\frac{1}{\left(2k\right)!}\,_{2}F_{1}\left(-k,-k+1/2;1;1\right) $$ where $\left (\right)_{j} $ is the Pochhammer symbol. Then, using the identity $$\,_{2}F_{1}\left(a,b;c;1\right)=\frac{\Gamma\left(c\right)\Gamma\left(c-a-b\right)}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)} $$ we have $$\sum_{j=0}^{k}\frac{4^{-j}}{j!^{2}\left(2k-2j\right)!}=\frac{1}{\left(2k\right)!}\frac{\Gamma\left(2k+1/2\right)}{\Gamma\left(k+1\right)\Gamma\left(k+1/2\right)}=\frac{4^{k}\Gamma\left(2k+1/2\right)}{\sqrt{\pi}\left(2k\right)!^{2}}=\frac{\left(4k\right)!}{4^{k}\left(2k\right)!^{3}} $$ by Gamma duplication formula. So $$\cos\left(x\right)J_{0}\left(x\right)=\sum_{k\geq0}\frac{\left(-1\right)^{k}\left(4k\right)!}{4^{k}\left(2k\right)!^{3}}x^{2k} $$ then your identity with $x=1 $. P. s. Me gustaría agradecer a Elajan para la inspiración.
Roger Hoover
Puntos
56
Esta respuesta se debe a Giulio Francot, un amigo mío. Su solución es extremadamente cercana a mis manipulaciones originales.
Tenemos: \begin{eqnarray*} \int\limits_{0}^{\frac{\pi }{2}}\sin ^{2}(\sin ^{2}(x))dx &=&\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (2\sin ^{2}(x))}{2}dx \\ &=&\frac{ \pi }{4}-\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}\sum\limits_{k\geq 0} \frac{(-1)^{k}4^{k}}{(2k)!}\sin ^{4k}(x)dx \\ &=&\frac{\pi }{4}-\frac{1}{2}\sum\limits_{k\geq 0}\frac{(-1)^{k}4^{k}}{(2k)! }\int\limits_{0}^{\frac{\pi }{2}}\sin ^{4k}(x)dx \\ &=&\frac{\pi }{4}-\frac{1}{2} \sum\limits_{k\geq 0}\frac{(-1)^{k}4^{k}}{(2k)!}\frac{(4k)!}{ (4^{k}(2k)!)^{2}}\frac{\pi }{2} \\ &=&\frac{\pi }{4}-\frac{\pi }{4}\sum\limits_{k\geq 0}\frac{(-1)^{k}}{ ((2k)!)^{3}}\frac{(4k)!}{4^{k}}.\tag{1} \end {eqnarray *}
Por otra parte:
\begin{eqnarray*} \int\limits_{0}^{\frac{\pi }{2}}\sin ^{2}(\sin ^{2}(x))dx &=&\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (2\sin ^{2}(x))}{2} dx=\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (1-\cos (2x))}{2}dx \\ &=&\frac{\pi }{4}-\int\limits_{0}^{\frac{\pi }{2}}\frac{\cos (1)\cos (\cos (2x))}{2}dx-\int\limits_{0}^{\frac{\pi }{2}}\frac{\sin (1)\sin (\cos (2x))}{ 2}dx \\ &=&\frac{\pi }{4}-\frac{\cos (1)}{2}\int\limits_{0}^{\frac{\pi }{2}}\cos (\cos (2x))dx=\frac{\pi }{4}-\frac{\cos (1)}{4}\int\limits_{0}^{\pi }\cos (\cos (x))dx \\ &=&\frac{\pi }{4}-\frac{\cos (1)\pi }{4}J_{0}(1).\tag{2} \end {eqnarray *}
