I'm trying to find the value of $$\int_0^\infty e^{-ax}\cos(bx)\,dx,\quad a>0.$$
by integrating $e^{−Az} ,A=\sqrt{a^2+b^2}$, over an appropriate sector with angle $ω$, with $cos ω=\dfrac{a}{A}$.
I'm seeing some solutions and I have the following questions:
1) It seems $ω= cos^{−1}(a/A)$ is strictly between 0 and $\pi/2$?
2) Why is it true that $\cos(\theta) \geq 1 - \dfrac{2\theta}{\pi} $? how to prove?
Can someone answer both questions with details?
Hint: Not sure what you're asking, but you can find an antiderivative of the integrand by considering $$\int e^{ax}\cos bx\;dx + i\int e^{ax}\sin bx\;dx\tag{1}$$ $$\int e^{ax}(\cos bx + i\sin bx)\;dx =\int e^{(a+bi)x}\; dx$$ $$= \frac{e^{(a+bi)x}}{a+bi} + C$$ $$=\frac{e^{ax}e^{ibx}(a-bi)}{(a+bi)(a-bi)} + C$$ $$=\frac{e^{ax}(\cos bx + i \sin bx)(a-bi)}{a^2+b^2} + C$$ $$=\left[\frac{e^{ax}}{a^2+b^2}\left(a\cos bx + b\sin bx\right)+C_1\right] + i\left[\frac{e^{ax}}{a^2+b^2}\left(a\sin bx - b\cos bx\right)+C_2\right]\tag{2}$$ Then equate real parts of (1) and (2) to get what you need to proceed with the definite integration. Is that helpful?