Improper integral of complex exponential

Question

I do not understand why the following holds

$$\int_{0}^{+\infty}e^{i \alpha x }dx=-\frac{1}{i \alpha } \,\,\,\,\, ,\,\,\,\, \alpha \in \mathbb{R}$$

I can see that

$$\int_{0}^{+\infty}e^{i \alpha x }dx=\lim_{c \to +\infty}\frac{1}{i \alpha }e^{i \alpha x}|_{0}^{c}=\lim_{c \to +\infty}\frac{1}{i \alpha }e^{i \alpha c}-\frac{1}{i \alpha }$$ But I do not see how

$$\lim_{c \to +\infty}\frac{1}{i \alpha }e^{i \alpha c}=0$$

Answer 1

The integral does not converge nor diverge. To give meaning to such an integral you must employ a regularization scheme. As an example multiply the integrand by $e^{-\epsilon x}$ with $\epsilon >0$ and at the end of the calculation let $\epsilon$ tend to 0

$$ \int_{0}^{+\infty}e^{(i \alpha -\epsilon)x }dx=\lim_{c \to +\infty}\frac{1}{i \alpha -\epsilon}e^{(i \alpha -\epsilon) x}|_{0}^{c}=\lim_{c \to +\infty}\frac{1}{i \alpha -\epsilon}e^{(i \alpha -\epsilon) c}-\frac{1}{i \alpha -\epsilon }=-\frac{1}{(i \alpha -\epsilon) } $$ $$ \lim_{\epsilon\to0}\int_{0}^{+\infty}e^{(i \alpha -\epsilon)x }dx=\lim_{\epsilon\to0}\frac{-1}{i \alpha -\epsilon }=\frac{-1}{i \alpha } $$ Another reference in the stack: https://physics.stackexchange.com/questions/7462/fourier-transform-of-the-coulomb-potential

Answer 2

The integral $\int_0^\infty e^{i\alpha x}\,dx$ fails to converge as an improper Riemann integral or a Lebesgue integral.

However, the symbol "$\displaystyle \int_0^\infty e^{i\alpha x}\,dx$" may be interpreted as the Fourier Transform of the Heaviside (unit step) function, which is a distribution (or generalized function).

In that case, we can write See here for an explanation

$${\int_0^\infty e^{i\alpha x}\,dx=\mathscr{F}\{H\}(\alpha)=\text{PV}\left(-\frac{1}{i\alpha}\right)+\pi \delta(\alpha)}$$

where $\delta(\alpha)$ is the Dirac Delta.

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Note that the regularization discussed in Francesco's answer, leads to

$$\lim_{\varepsilon\to 0^+} \int_0^\infty e^{i(\alpha +i\varepsilon)x}\,dx=\lim_{\varepsilon\to 0^+} \frac{1}{\varepsilon-i\alpha} \tag 1$$

And in THIS ANSWER, I showed that the limit, $\lim_{y\to 0^+}\frac{1}{x+iy}$, is given in distribution by

$$\lim_{y\to 0^+}\frac{1}{x+iy}\sim \text{PV}\left(\frac1x\right)-i\pi \delta(x)\tag 2$$

where $\delta(x)$ it the Dirac Delta.

Applying the result in $(2)$ to $(1)$ reveals

$$\bbox[5px,border:2px solid #C0A000]{\lim_{\varepsilon\to 0^+} \int_0^\infty e^{i(\alpha +i\varepsilon)x}\,dx\sim \text{PV}\left(-\frac{1}{i\alpha}\right)+\pi \delta(\alpha)}$$

as expected!