Suppose that I need to express through Fourier integral the following function
$$f(t)=\begin{cases} 0 & t< 0 \\ C e^{-(\gamma+i\omega)t} & t\geq 0 &(C,\omega,\gamma \in\mathbb{R})\end{cases}$$
For $t<0$ , $f(t)=0$ is ok, while for $t\geq0$ my textbook report the following:
$$f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} K(\omega')e^{-i\omega't} \mathrm{d}\omega'$$
Where $ K(\omega')$ is the Fourier Trasform:
$$K(\omega')=\frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty} f(t)e^{+i\omega't} \mathrm{d}t\tag{A}$$
The integral starts from $t=0$. I do not undertstand this choice of signes since usually a Fourier trasform is defined with a complex exponential that has a $-$ sign in the exponent, while here there is a $+$ sign. Is this related to the fact that usually the Fourier trasform is defined using $k$ (the wavenumber) while here it is used with $\omega$ (the frequency)? This does not seem the answer, since, for example, here the signes are opposite to the one used.
Using the above expression anyway:
$$K(\omega')=\frac{C}{\sqrt{2\pi}}\int_{0}^{+\infty} e^{-(\gamma+i\omega)t} e^{+i\omega't} \mathrm{d}t=\frac{C}{\sqrt{2\pi}}\int_{0}^{+\infty} e^{[i(\omega'-\omega)-\gamma]t} \mathrm{d}t$$
On textbook it is claimed that
$$\frac{C}{\sqrt{2\pi}}\int_{0}^{+\infty} e^{[i(\omega'-\omega)-\gamma]t} \mathrm{d}t=\frac{C}{\sqrt{2\pi}} \frac{-1}{i(\omega'-\omega)-\gamma}\tag{B}$$
But I really do not see why this integral gives that result, since I think that it is divergent.
To sum up my questions are:
- (most important) How is equality in $(B)$ true?
- How is $(A)$ a possible form of Fourier trasfrom? Is it possible to have a trasform that starts from $t=0$ and not $t=-\infty$? Also, what about that use of signes in the exponentials?
First, Eq. $(B)$ is true only when $\gamma > 0$. You can verify that by using $$\int_{a}^{b} e^{\lambda x} dx = \frac{1}{\lambda}\left(e^{\lambda b} - e^{\lambda a}\right).$$
Regarding Eq. $(A)$, the limit starts from $0$ because $f(t)=0$ for $t <0$. In particular, if $g(t)=0$ for $t < a$ then $$\int_{-\infty}^{+\infty} g(t) dt = \int_{-\infty}^{a} g(t) dt + \int_{a}^{+\infty} g(t) dt = \int_{a}^{+\infty} g(t) dt.$$
Last, if you use the standard definitions of Fourier transform, i.e., $$f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(w') e^{i w' t} dw',$$ and $$F(w') = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(t) e^{-i w' t} dt,$$ then $K'(w') = F(-w')$.