一个求解积分的问题

I=\displaystyle \int_{0}^{\pi}\frac{\cos mx}{5-4\cos x}dx\\{}\\ m\in Z^+

\displaystyle I=\frac{1}{2}\int_{-\pi}^{\pi}\frac{\cos mx}{5-4\cos x}dx\\{}\\ set \ I_1=\int_{-\pi}^{\pi}\frac{\cos mx}{5-4\cos x}dx,\ I_2=\int_{-\pi}^{\pi}\frac{\cos mx}{5-4\cos x}dx\\{}\\ \therefore\ I_1+iI_2=\int_{-\pi}^{\pi}\frac{e^{imx}}{5-4\cos x}dx\\{}\\ set\ z=e^{ix}\\{}\\ \therefore\ I_1+iI_2=\frac{1}{i}\int_\Gamma\frac{z^m}{5z-2(1+z^2)}dz=\frac{i}{2}\int_\Gamma\frac{z^m}{\left(z-\frac{1}{2}\right)(z-2)}dz\\{}\\ within\ \Gamma,\ f(z)\ just\ only\ one\ order\ 1\ pole\ z=\frac{1}{2}\\{}\\ \mathop{Res}\limits_{z=\frac{1}{2}}f(z)=\frac{z^m}{z-2}\bigg|_{z=\frac{1}{2}}=-\frac{1}{3\cdot2^{m-1}},\\{}\\ \because Residue\ theorem\ I_1+iI_2=\left(-\frac{1}{2i}\right)\cdot2\pi i\left(-\frac{1}{3\cdot2^{m-1}}\right)=\frac{\pi}{3\cdot2^{m-1}},\\{}\\ I_1=\frac{\pi}{3\cdot2^{m-1}},\ I_2=0,\ \therefore\ I=\frac{1}{2}I_1=\frac{\pi}{3\cdot2^m}

\displaystyle \int_0^\infty\frac{\sin x}{x}dx\ (有阻尼的震动)\\{}\\ \int_0^\infty\sin x^2dx\ (光的折射)\\{}\\ \int_0^\infty e^{-ax^2}\cos bxdx(a>0)\ (热传导)\\{}\\ \cdots\cdots等等

7个月前 评论
thus 7个月前

\displaystyle I=\frac{1}{2}\int_{-\pi}^{\pi}\frac{\cos mx}{5-4\cos x}dx\\{}\\ set \ I_1=\int_{-\pi}^{\pi}\frac{\cos mx}{5-4\cos x}dx,\ I_2=\int_{-\pi}^{\pi}\frac{\cos mx}{5-4\cos x}dx\\{}\\ \therefore\ I_1+iI_2=\int_{-\pi}^{\pi}\frac{e^{imx}}{5-4\cos x}dx\\{}\\ set\ z=e^{ix}\\{}\\ \therefore\ I_1+iI_2=\frac{1}{i}\int_\Gamma\frac{z^m}{5z-2(1+z^2)}dz=\frac{i}{2}\int_\Gamma\frac{z^m}{\left(z-\frac{1}{2}\right)(z-2)}dz\\{}\\ within\ \Gamma,\ f(z)\ just\ only\ one\ order\ 1\ pole\ z=\frac{1}{2}\\{}\\ \mathop{Res}\limits_{z=\frac{1}{2}}f(z)=\frac{z^m}{z-2}\bigg|_{z=\frac{1}{2}}=-\frac{1}{3\cdot2^{m-1}},\\{}\\ \because Residue\ theorem\ I_1+iI_2=\left(-\frac{1}{2i}\right)\cdot2\pi i\left(-\frac{1}{3\cdot2^{m-1}}\right)=\frac{\pi}{3\cdot2^{m-1}},\\{}\\ I_1=\frac{\pi}{3\cdot2^{m-1}},\ I_2=0,\ \therefore\ I=\frac{1}{2}I_1=\frac{\pi}{3\cdot2^m}

\displaystyle \int_0^\infty\frac{\sin x}{x}dx\ (有阻尼的震动)\\{}\\ \int_0^\infty\sin x^2dx\ (光的折射)\\{}\\ \int_0^\infty e^{-ax^2}\cos bxdx(a>0)\ (热传导)\\{}\\ \cdots\cdots等等

7个月前 评论
thus 7个月前