一道反常积分的计算题

计算$$\int_0^{+\infty} \dfrac{1}{1+x^6}\,dx$$

解答:本题可以利用一个结论

$$\int_0^{+\infty}\dfrac{x^{p-1}}{1+x}\,dx=\Gamma(p)\Gamma(1-p)=\dfrac{\pi}{\sin(\pi p)},\quad 0<p<1$$

因此

$$\int_0^{+\infty} \dfrac{1}{1+x^6}\,dx=\dfrac{1}{6}\int_0^{+\infty} \dfrac{t^{\frac{1}{6}-1}}{1+t}\,dt(t=x^6)=\dfrac{1}{6}\Gamma(\dfrac{1}{6})\Gamma(\dfrac{5}{6})=\dfrac{1}{6}\times\dfrac{\pi}{\sin(\dfrac{\pi}{6})}=\dfrac{\pi}{3}$$

 

同时也可以用留数来解答:函数$\dfrac{1}{1+{{x}^{6}}}$有6个一阶极点：${{a}_{k}}={{e}^{\frac{\left( 2k+1 \right)\pi }{6}i}}$. 当$k=0,1,2$时，$\operatorname{Im}{{a}_{k}}>0$.

所以有
$$\int_{0}^{+\infty }{\frac{1}{1+{{x}^{6}}}dx}=\frac{1}{2}\int_{-\infty }^{+\infty }{\frac{1}{1+{{x}^{6}}}dx}=\pi i\sum\limits_{\operatorname{Im}{{a}_{k}}>0}^{{}}{\underset{z={{a}_{k}}}{\mathop{\operatorname{Res}}}\,\frac{1}{1+{{x}^{6}}}}
=\pi i\sum\limits_{k=0}^{2}{\underset{z={{a}_{k}}}{\mathop{\operatorname{Res}}}\,\frac{1}{1+{{x}^{6}}}}=\pi i\left( \frac{1}{6a_{0}^{5}}+\frac{1}{6a_{1}^{5}}+\frac{1}{6a_{2}^{5}} \right)==\frac{\pi i}{6}\left( -{{a}_{0}}-{{a}_{1}}-{{a}_{2}} \right)=\frac{\pi }{3}$$