How to integral \(\int_0^\infty \exp(-(a+bi)x)dx\)?
Method 1: use Euler's formula:
\[\int_0^{\infty}e^{-(a+bi)x}dx=\int_0^{\infty}e^{-ax}(\cos(-bx)+\sin(-bx)i)dx\\ =\int_0^{\infty}e^{-ax}\cos(-bx)dx + i\int_0^{\infty}e^{-ax}\sin(-bx)dx\\ =\frac{a}{a^2+b^2}+i(-\frac{b}{a^2+b^2})=\frac{a-bi}{a^2+b^2} \]Method 2: Simple approach
\[\int_0^{\infty}e^{-(a+bi)x}dx=\frac{1}{a+bi}\int_0^{\infty}e^{-(a+bi)x}d(a+bi)x=\frac{1}{a+bi}\int_0^{\infty}e^{-t}dt=\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2} \] 标签：infty,frac,int,bi,dx,bx From： https://www.cnblogs.com/uceec00/p/16984530.html