27024

27024 (Gheorghe Szöllösy)

Fie $I_{n}=\int _{0}^{\pi }{\frac {\cos nx}{13-12\cos x}}\,dx,n\geq 0.$ Să se calculeze $\lim _{n\to \infty }(I_{0}+I_{1}+I_{2}+\ldots +I_{n}).$

Soluție. Să observăm că

$$I_{n+2}+I_{n}=\int _{0}^{\pi }{\frac {2\cos x-\cos(n_{1})x}{13-12\cos x}}\,dx={\frac {1}{6}}\int _{0}^{\pi }{\frac {(12\cos x-13+13)\cos(n+1)x}{13-12\cos x}}\,dx=$$

$$={\frac {1}{6(n+1)}}\sin(n+1)x{\Biggr |}_{0}^{\pi }+{\frac {13}{6}}I_{n+1},$$

oricare ar fi$n\in \mathbb {N} .$ Atunci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_n=\alpha\left(\frac{2}{3}\right)^n+\beta\left(\frac{3}{2}\right)^n} , unde Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha+\beta=I_0=\frac{\pi}{5}} și ${\frac {2}{3}}\alpha +{\frac {3}{2}}\beta =I_{1}={\frac {2\pi }{15}}.$

Obținem $\alpha ={\frac {\pi }{5}},\beta =0$ și $I_{n}={\frac {\pi }{5}}\left({\frac {2}{3}}\right)^{n}$ pentru orice $n>0$.

În consecință,

$$\lim _{n\to \infty }(I_{0}+I_{1}+I_{2}+\ldots +I_{n})={\frac {\frac {\pi }{5}}{1-{\frac {2}{3}}}}={\frac {3\pi }{5}}.$$