2021年考研数二第11题解析：无穷限定积分的计算、极限的计算

一、题目

难度评级：

二、解析

$$
\begin{aligned}
\int_{-\infty}^{+\infty}\left| x \right|3^{-x^{2}}\mathrm{~d}x & = 2 \int_{0}^{+\infty} x 3^{-x^{2}}\mathrm{~d}x \\ \\
& = -\int_{0}^{+\infty}3^{-x^{2}}\mathrm{~d}\left(-x^{2}\right) \\ \\
& = \textcolor{gray}{ t = -x^{2}, x \in \left( 0, + \infty \right) \leadsto t \in \left( 0, – \infty \right) } \\ \\
& = – \int_{0}^{- \infty} 3^{t} \mathrm{~d} t \\ \\
& = \int_{- \infty}^{0} 3^{t} \mathrm{~d} t
\end{aligned}
$$

又因为：

$$
\left( 3^{t} \right) ^{\prime} = 3^{t} \ln 3
$$

所以：

$$
\int 3^{t} \mathrm{~d} t = \frac{1}{\ln 3} \cdot 3^{t}
$$

于是：

$$
\begin{aligned}
\int_{-\infty}^{+\infty}\left| x \right|3^{-x^{2}}\mathrm{~d}x & = \int_{- \infty}^{0} 3^{t} \mathrm{~d} t \\ \\
& = \left. \frac{1}{\ln 3} \cdot 3^{t} \right|_{- \infty}^{0} \\ \\
& = \frac{1}{\ln 3} \cdot 1 – 0 \\ \\
& = \frac{1}{\ln 3}
\end{aligned}
$$

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