求(sin(πax) e^(-iπux)) / πx的积分
时间: 2023-05-29 22:03:48 浏览: 48
这是一个复合函数积分,需要使用分部积分法。设$f(x)=\frac{1}{\pi x}$,$g(x)=\sin(\pi ax)e^{-i\pi ux}$,则有:
$$\int \frac{\sin(\pi ax)e^{-i\pi ux}}{\pi x}dx=\int f(x)g(x)dx$$
对 $f(x)$ 求导得:
$$f'(x)=-\frac{1}{\pi x^2}$$
对 $g(x)$ 积分得:
$$\int g(x)dx=-\frac{1}{\pi u}e^{-i\pi ux}$$
根据分部积分公式,有:
$$\int f(x)g(x)dx=f(x)\int g(x)dx-\int f'(x)\int g(x)dx dx$$
代入上面的式子得:
$$\int \frac{\sin(\pi ax)e^{-i\pi ux}}{\pi x}dx=-\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi u x}-\frac{a}{u}\int\frac{\cos(\pi ax)e^{-i\pi ux}}{x}dx$$
这个积分还不能直接求解,需要再次分部积分,设$h(x)=\frac{\cos(\pi ax)e^{-i\pi ux}}{x}$,则有:
$$h'(x)=-\frac{\pi a\sin(\pi ax)e^{-i\pi ux}}{x}-i\pi u\frac{\cos(\pi ax)e^{-i\pi ux}}{x}+\frac{\cos(\pi ax)e^{-i\pi ux}}{x^2}$$
$$\int h(x)dx=\int\frac{\cos(\pi ax)e^{-i\pi ux}}{x}dx=\cos(\pi ax)e^{-i\pi ux}\ln(x)+i\pi a\int\frac{\sin(\pi ax)e^{-i\pi ux}}{x}dx-\int\frac{\cos(\pi ax)e^{-i\pi ux}}{x^2}dx$$
代入上面的式子得:
$$\int \frac{\sin(\pi ax)e^{-i\pi ux}}{\pi x}dx=-\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi u x}-\frac{a}{u}\left[\cos(\pi ax)e^{-i\pi ux}\ln(x)-i\pi a\int\frac{\sin(\pi ax)e^{-i\pi ux}}{x}dx+\int\frac{\cos(\pi ax)e^{-i\pi ux}}{x^2}dx\right]$$
移项得:
$$\left(1+\frac{a^2}{u^2}\right)\int\frac{\sin(\pi ax)e^{-i\pi ux}}{x}dx=-\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi u x}+\frac{a}{u}\cos(\pi ax)e^{-i\pi ux}\ln(x)+\frac{a}{u^2}\int\frac{\cos(\pi ax)e^{-i\pi ux}}{x^2}dx$$
将第二个积分式子再次代入上面的式子得:
$$\left(1+\frac{a^2}{u^2}\right)\int\frac{\sin(\pi ax)e^{-i\pi ux}}{x}dx=-\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi u x}+\frac{a}{u}\cos(\pi ax)e^{-i\pi ux}\ln(x)+\frac{a^2}{u^2}\int\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi x}dx$$
移项得:
$$\left(1-\frac{a^2}{u^2}\right)\int\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi x}dx=\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi u x}-\frac{a}{u}\cos(\pi ax)e^{-i\pi ux}\ln(x)$$
最终得到:
$$\int\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi x}dx=\frac{\sin(\pi ax)e^{-i\pi ux}}{\pi u x\left(1-\frac{a^2}{u^2}\right)}-\frac{a}{u}\frac{\cos(\pi ax)e^{-i\pi ux}}{\pi\left(1-\frac{a^2}{u^2}\right)}\ln(x)+C$$
其中 $C$ 是常数。