微积分(A)每日一题[41]

判断敛散性：\(\int_0^{+\infty} \frac{\sin^2x}{x^p}dx\)

\[\begin{aligned} &当 p \le 0 时显然发散 \\ &\int_0^{+\infty} \frac{\sin^2x}{x^p}dx= \int_0^{1} \frac{\sin^2x}{x^p}dx+ \int_1^{+\infty} \frac{\sin^2x}{x^p}dx \\ &\lim_{x \to 0^+}x^{p-2} \frac{\sin^2x}{x^p}=1 \Rightarrow p-2<1 \Rightarrow p<3时收敛(0<p<3) \\ &\lim_{x \to +\infty}x^{r} \frac{\sin^2x}{x^p}=l \Rightarrow p-r>0 \land r>1 \Rightarrow p>1 \\ &故 1<p<3时收敛 \end{aligned} \]

标签：infty,frac,int,微积分,2x,41,每日,dx,sin
来源： https://www.cnblogs.com/nekko/p/15728415.html