反常积分例题
Instance 0
\[\begin{align}
\int_{-\infty}^{+\infty}\frac{1}{1+x^{2}}dx=?
\\ \\
\int_{-\infty}^{+\infty}\frac{1}{1+x^{2}}=\int_{-\infty}^{0}\frac{1}{1+x^{2}}dx+\int_{0}^{+\infty}\frac{1}{1+x^{2}}dx
\\ \\ \\ \\
\int_{-\infty}^{0}\frac{1}{1+x^{2}}dx=\lim_{a\to\infty}\int_{a}^{0}\frac{1}{1+x^{2}}dx
\\ \\
\lim_{a\to -\infty}\int_{a}^{0}\frac{1}{1+x^{2}}dx=\lim_{a\to-\infty}\left[\arctan{(x)}\right]_{a}^{0}
\\ \\ \\ \\
\int_{0}^{+\infty}\frac{1}{1+x^{2}}dx=\lim_{b\to+\infty}\int_{0}^{b}\frac{1}{1tx^{2}}dx
\\ \\
\lim_{b\to+\infty}\int_{0}^{b}\frac{1}{1+x^{2}}dx=\lim_{b\to+\infty}\left[\arctan{(x)}\right]_{0}^{b}
\\ \\ \\ \\
\text{arctan}函数的值域为: (-\frac{\pi}{2},\frac{\pi}{2})
\\ \\
\therefore\lim_{a\to-\infty}\arctan(a)=-\frac{\pi}{2}
\\ \\
\lim_{b\to+\infty}\arctan(b)=\frac{\pi}{2}
\\ \\
此外:
\arctan(0)=0
\\ \\
[0-(-\frac{\pi}{2})]+[\frac{\pi}{2}-0]=\pi
\\ \\
\Rightarrow\int_{-\infty}^{+\infty}\frac{1}{1+x^{2}}dx=\pi
\end{align}
\]
