$\int_0^{\infty}\frac{f(x,2m-1)-\sin x}{x^{2m+1}}dx$其中，f(x,2m-1)表示sinx的2m-1阶泰勒展开

$$\int_0^{\infty}\frac{x-\sin x}{x^3}dx$$
m=2时
$$\int_0^{\infty}\frac{x-\frac{x^3}{6}-\sin x}{x^5}dx$$

$\frac{\pi(-1)^{m-1}}{2(2m)!}$

$$\int_0^{\infty} \frac{\sin x}{x}dx$$

$$G(a)=\int_{m(a)}^{n(a)} f(x,a)dx$$

$$G(a)=F(n(a),a)-F(m(a),a)$$

$$\int \frac{1}{\cos^3 \theta} d\theta$$

plane circular restricted three-body problem
02.04有重要修正！！

The Three Body Problem and its Classical Integration