I simplified an integral to $$\int_{0}^{\infty}{e^{(i\xi-a)x}dx},$$ where $i$ is the imaginary number, $\xi$, and $a\gt0$ are constants. My textbook says this integral evaluates to $$\frac{1}{a-i\xi}$$
I tried doing this $$\int_{0}^{\infty}{e^{(i\xi-a)x}dx}=\int_{0}^{\infty}{e^{-ax}e^{i\xi x}dx}=$$ $$\int_{0}^{\infty}{e^{-ax}[\cos{(\xi x)}+i\sin{(\xi x)}]dx}=$$ $$\int_{0}^{\infty}{e^{-ax}\cos{(\xi x)}dx}+i\int_{0}^{\infty}{e^{-ax}\sin{(\xi x)}dx}=$$ $$\frac{a}{a^2+\xi^2}+\frac{i\xi}{a^2+\xi^2}$$
Does anyone have an idea of how my textbook got $\frac{1}{a-i\xi}$?
Just multiply the numerator and denominator by conjugate of denominator:
$$\dfrac{1}{a-i\xi}\times\dfrac{a+i\xi}{a+i\xi}$$
$$=\dfrac{a}{a^2+\xi^2} + \dfrac{i\xi}{a^2+\xi^2}$$