一、题目描述
$$
\lim_{x \rightarrow + \infty} \frac{(1+\frac{1}{x})^{x^{2}}}{e^{x}} = ?
$$
$$
\lim_{x \rightarrow + \infty} \frac{(1+\frac{1}{x})^{x^{2}}}{e^{x}} = ?
$$
因为对于 $\sqrt[3]{x^{2}}$ 而言,必须有 $x$ $\neq$ $0$, 于是,在区间 $[-1, 1]$ 内,定积分 $\int_{-1}^{1}$ $\frac{1}{\sqrt[3]{x^{2}}}$ $\mathrm{d} x$ 其实是一个瑕积分,瑕点就是 $x$ $=$ $0$, 由于在真正进行积分运算的时候,被积函数不能包含瑕点,所以,我们必须在 $x$ $=$ $0$ 处对原积分进行“分割”。
函数 $y$ $=$ $\frac{1}{\sqrt[3]{x^{2}}}$ 的示意图像如下:
直接来看,这是一个上限趋于无穷的的反常积分,但其实,由于被积函数中的 $\sqrt{x}$ 必须有 $x$ $>$ $0$, 因此,该反常积分的下限也需要通过取极限的方式才能在计算中使用:
我们可以引入两个变量 $s$ 和 $t$, 并使 $s$ $\rightarrow$ $0^{+}$, $t$ $\rightarrow$ $\infty$, 以此来代替该反常积分原来的上限和下限。
同时,由于 $1$ 具有 $1^{2}$ $=$ $1$ 等特殊性质,因此,我们将 $1$ 作为分割区间 $[0, \infty]$ 的一个中间点。
继续阅读“反常积分 $\int_{0}^{\infty}$ $\frac{1}{(1 + x)\sqrt{x}}$ $\mathrm{d} x$ 的计算方法”$$\int_{\textcolor{Orange}{a}}^{\textcolor{Orange}{b}} \textcolor{Red}{u}(x) \mathrm{d} x =$$ $$\int_{\textcolor{Orange}{a}}^{\textcolor{Orange}{b}} \textcolor{Red}{u}(x) \textcolor{Green}{\cdot} \textcolor{Red}{x} ^{\textcolor{Yellow}{\prime}} \mathrm{d} x =$$ $$\textcolor{Red}{x} \textcolor{Green}{\cdot} \textcolor{Red}{u}(x) \Bigg|_{\textcolor{Orange}{a}}^{\textcolor{Orange}{b}}$$ $$\textcolor{Green}{-}$$ $$\int_{\textcolor{Orange}{a}}^{\textcolor{Orange}{b}} \textcolor{Red}{x} \textcolor{Green}{\cdot} \textcolor{Red}{u} ^{\textcolor{Yellow}{\prime}} (x) \mathrm{d} x.$$
$$\int_{\textcolor{Orange}{a}}^{\textcolor{Orange}{b}} \Bigg[ \textcolor{Red}{F}(x) \textcolor{Green}{\cdot} \textcolor{Red}{M} ^{\textcolor{Yellow}{\prime}}(x) \Bigg] \mathrm{d} x =$$ $$\Bigg[ \textcolor{Red}{F}(x) \textcolor{Green}{\cdot} \textcolor{Red}{M}(x) \Bigg] \Bigg|_{\textcolor{Orange}{a}}^{\textcolor{Orange}{b}}$$ $$\textcolor{Green}{-}$$ $$\int_{\textcolor{Orange}{a}}^{\textcolor{Orange}{b}} \Bigg[ \textcolor{Red}{F} ^{\textcolor{Yellow}{\prime}}(x) \textcolor{Green}{\cdot} \textcolor{Red}{M}(x) \Bigg] \mathrm{d} x.$$
则:以下使用函数 $\textcolor{Orange}{\phi(t)}$ 对定积分 $\textcolor{Orange}{\int_{a}^{b}}$ $\textcolor{Orange}{f(x)}$ $\textcolor{Orange}{\mathrm{d} x}$ 进行换元的选项中,正确的是哪个?
$$\int_{\textcolor{Yellow}{a}}^{\textcolor{Yellow}{b}} \textcolor{Red}{f}(\textcolor{Red}{x}) \mathrm{d} x =$$ $$\int_{\textcolor{Yellow}{\alpha}}^{\textcolor{Yellow}{\beta}} \textcolor{Red}{f} [\textcolor{Red}{\phi}(\textcolor{Red}{t})] \mathrm{d} [\textcolor{Red}{\phi} (\textcolor{Red}{t})] =$$ $$\int_{\textcolor{Yellow}{\alpha}}^{\textcolor{Yellow}{\beta}} \textcolor{Red}{f} [\textcolor{Red}{\phi}(\textcolor{Red}{t})] \textcolor{Green}{\cdot} \textcolor{Red}{\phi} ^{\textcolor{Orange}{\prime}} (\textcolor{Red}{t}) \mathrm{d} t.$$注意:对定积分的换元,不仅要更换自变量 $x$, 还要对应的更换积分上下限.
$$\int_{\textcolor{Orange}{0}}^{\textcolor{Orange}{\frac{\pi}{2}}} \textcolor{Yellow}{\sin} ^{\textcolor{Red}{n}} \textcolor{Yellow}{x} \mathrm{d} x =$$ $$\int_{\textcolor{Orange}{0}}^{\textcolor{Orange}{\frac{\pi}{2}}} \textcolor{Yellow}{\cos} ^{\textcolor{Red}{n}} \textcolor{Yellow}{x} \mathrm{d} x =$$ $$\frac{\textcolor{Red}{n-1}}{\textcolor{Red}{n}} \cdot \frac{\textcolor{Red}{n-3}}{\textcolor{Red}{n-2}} \cdots \frac{\textcolor{Red}{2}}{\textcolor{Red}{3}} \cdot \textcolor{Red}{1}.$$
$$\int_{\textcolor{Orange}{0}}^{\textcolor{Orange}{\frac{\pi}{2}}} \textcolor{Yellow}{\sin} ^{\textcolor{Red}{n}} \textcolor{Yellow}{x} \mathrm{d} x =$$ $$\int_{\textcolor{Orange}{0}}^{\textcolor{Orange}{\frac{\pi}{2}}} \textcolor{Yellow}{\cos} ^{\textcolor{Red}{n}} \textcolor{Yellow}{x} \mathrm{d} x =$$ $$\frac{\textcolor{Red}{n-1}}{\textcolor{Red}{n}} \cdot \frac{\textcolor{Red}{n-3}}{\textcolor{Red}{n-2}} \cdots \frac{\textcolor{Red}{1}}{\textcolor{Red}{2}} \cdot \frac{\textcolor{Red}{\pi}}{\textcolor{Red}{2}}.$$
$$\int_{\textcolor{Red}{a}}^{\textcolor{Red}{a + T}} f(x) \mathrm{d} x =$$ $$\int_{\textcolor{Red}{0}}^{\textcolor{Red}{T}} f(x) \mathrel{d} x =$$ $$\int_{\textcolor{Red}{-\frac{T}{2}}}^{\textcolor{Red}{\frac{T}{2}}} f(x) \mathrel{d} x.$$ 注意:对于被积函数是同一个周期函数的定积分而言,只要上限与下限的差值相等,则这两个定积分就是相等的.
此外:
$$\int_{\textcolor{Red}{0}}^{\textcolor{Red}{a \cdot T}} f(x) \mathrm{d} x =$$ $$\textcolor{Orange}{a} \textcolor{Green}{\cdot} \int_{\textcolor{Red}{0}}^{\textcolor{Red}{T}} f(x) \mathrm{d} x.$$
$$F^{\textcolor{Yellow}{\prime}} =$$ $$\Bigg[ \int_{\textcolor{Orange}{\mu(x)}}^{\textcolor{Orange}{\phi(x)}} f(t) \mathrm{d} t \Bigg]^{\textcolor{Yellow}{\prime}} =$$ $$f[\textcolor{Orange}{\phi(x)}] \textcolor{Green}{\cdot} \textcolor{Orange}{\phi} ^{\textcolor{Yellow}{\prime}} \textcolor{Orange}{(x)}$$ $$\textcolor{Green}{-}$$ $$f[\textcolor{Orange}{\mu(x)}] \textcolor{Green}{\cdot} \textcolor{Orange}{\mu} ^{\textcolor{Yellow}{\prime}} \textcolor{Orange}{(x)}.$$