How to show $\int_0^\infty \cos\left(-1+\frac{ix}{\omega}\right)e^{-x}dx $ converges with epsilon-delta proof? I have some integrals like 
$$\int_0^\infty \cos\left(-1+\frac{ix}{\omega}\right)e^{-x}dx,$$
and  I want to show that they exist with epsilon-delta definition. Can you give me any tips about it? I already showed convergence with limit but I need the epsilon-delta proof.
 A: Let $$f(t) = \int_0^t \cos\left(-1+\frac{ix}{\omega}\right)e^{-x}dx,$$
Let $\alpha = -1+\frac{ix}{\omega}$. We know that 
$$\cos\left(-1+\frac{ix}{\omega}\right)e^{-x} = \cos(\alpha)e^{-x} = \frac{e^{i\alpha } + e^{-i\alpha }}{2}e^{-x} = \frac{1}{2}e^{-i}e^{\frac{-x}{\omega} - x} + \frac{1}{2}e^{i}e^{\frac{x}{\omega} - x}$$
So we have $$f(t) = \frac{1}{2}e^{-i}\int_0^t e^{\frac{-x}{\omega} - x}dx + \frac{1}{2}e^{i}\int_0^t e^{\frac{x}{\omega} - x}dx = \frac{1}{2}e^{-i}\frac{1-e^{\frac{-t}{\omega} - t}}{\frac{1}{\omega} + 1} + \frac{1}{2}e^{i}\frac{e^{\frac{t}{\omega} - t} - 1}{\frac{1}{\omega} - 1} = \frac{-e^{\frac{-t}{\omega} - t - i}}{2(\frac{1}{\omega} + 1)} + \frac{e^{\frac{t}{\omega} - t + i}}{2 (\frac{1}{\omega} - 1)} + \frac{e^{-i}}{2 (\frac{1}{\omega} + 1)} - \frac{e^i}{2 (\frac{1}{\omega} - 1)}$$
Note that the two last terms are constant. If $\omega \gt 1$ or $\omega \lt -1$
$$\lim_{t \to \infty}\frac{-e^{\frac{-t}{\omega} - t - i}}{2(\frac{1}{\omega} + 1)} = \lim_{t \to \infty} \frac{e^{\frac{t}{\omega} - t + i}}{2 (\frac{1}{\omega} - 1)} =0$$ And $$\lim_{t \to \infty} f(t) = \frac{e^{-i}}{2 (\frac{1}{\omega} + 1)} - \frac{e^i}{2 (\frac{1}{\omega} - 1)}$$
When $-1\lt\omega \lt 1$ we have $\frac{-1}{\omega} - 1 \gt 1$ or $\frac{1}{\omega} - 1 \gt 1$ and so the improper integral diverges. Additionally $\omega = 1$ or $\omega = -1$ leads to a divergent improper integral because in those cases $f(t) = t + (\dots)$. If you want to use epsilon-delta definition, prove that $\lim_{x \to \infty} e^{ax} = 0$ if $a\lt 0$ and if $a \gt 0$ then $\lim_{x \to \infty} e^{ax} = \infty$.
Another way to do that is Laplace transform. We have $$\cos\left(-1+\frac{ix}{\omega}\right) = \cos(1) \cosh(\frac{x}{\omega}) + i \sin(1) \sinh(\frac{x}{\omega})$$ It's known that if $\Re\{s\} \gt |\alpha|$ $$\mathcal L \{\sinh \alpha x \} = \frac{\alpha}{s^2 - \alpha^2}$$
And 
$$\mathcal L \{\cosh \alpha x \} = \frac{s}{s^2 - \alpha^2}$$
Let $s = 1$ and $\alpha = \frac{1}{\omega}$ so if $1 \gt |\frac{1}{\omega}|$ or equivalently $|\omega| \gt 1$
$$\int_0^\infty \cos\left(-1+\frac{ix}{\omega}\right)e^{-x}dx = \cos(1)\frac{1}{1^2 - (\frac{1}{\omega})^2} + i \sin(1) \frac{\frac{1}{\omega}}{1^2 - (\frac{1}{\omega})^2}$$
which is the same as the former result after rearranging.
