- $$L=\lim_{n\rightarrow \infty }\int_{0}^{1}\dfrac{\ln(1+x^{n+k})}{\ln(1+x^{n})}dx$$
- $$\lim_{n\rightarrow \infty }n\left (\int_{0}^{1}\dfrac{\ln(1+x^{n+k})}{\ln(1+x^{n})}dx-L \right )$$
Cho $k\in \mathbb{N}^*$. Tìm $$L=\lim_{n\to \infty }\int_0^1 \dfrac{\ln(1+x^{n+k})}{\ln(1+x^{n})}\mathrm d x$$
Bắt đầu bởi Crystal , 04-01-2012 - 17:50
#2
Đã gửi 20-11-2015 - 22:03
$\int_{0}^{1}\frac{\ln{(1+x^{n+k})}}{\ln{(1+x^{n})}}=\int_{0}^{1-\varepsilon }\frac{\ln{(1+x^{n+k})}}{\ln{(1+x^{n})}}+\int_{1-\varepsilon }^{1}\frac{\ln{(1+x^{n+k})}}{\ln{(1+x^{n})}} =I_{1}+I_{2} $
$I_{2}=\varepsilon \frac{\ln{(1+{c}^{n+k})}}{\ln{(1+{c}^{n})}}=0 I_{1}=\int_{0}^{1-\varepsilon }\frac{x^{n+k}}{x^{n}}=\int_{0}^{1-\varepsilon }x^{k}$
$=\lim_{\varepsilon \to\ 0}\frac{x^{k+1}}{k+1}|^{1-\varepsilon }_{0} =\frac{1}{k+1}$
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