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

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

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    $\begingroup$ It's not true. The modulus of the integrand is $1$. The integral can never converge. $\endgroup$
    – Cauchy
    Aug 17, 2017 at 16:17
  • $\begingroup$ Maybe you are mixing this up with the Laplace transform of a constant function, which is $1/s$ for $\endgroup$
    – MrYouMath
    Aug 17, 2017 at 16:33
  • $\begingroup$ @Cauchy The modulus of $e^{ix^2}$ is $1$ and the integral converges. $\endgroup$
    – zhw.
    Aug 17, 2017 at 16:42
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    $\begingroup$ Read my answer, perhaps in your context that is what they use. $\endgroup$
    – Frank
    Aug 17, 2017 at 16:52

2 Answers 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.

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!

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    $\begingroup$ @soren The accepted answer is simply not correct. $\endgroup$
    – Mark Viola
    Mar 5, 2021 at 17:14

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

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    $\begingroup$ I'm willing to bet this is what was meant by the OP. $\endgroup$ Aug 17, 2017 at 17:06
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    $\begingroup$ And this is incomplete in the context of Generalized functions. $\endgroup$
    – Mark Viola
    Aug 17, 2017 at 17:09
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    $\begingroup$ But the inverse Fourier Transform of your result does not equal the Heaviside function. Rather, $$\mathscr{F}^{-1}\left(-\frac{1}{i\alpha}\right)=\frac12\text{sgn}(x)=H(x)+\frac12 \ne H(x)$$ $\endgroup$
    – Mark Viola
    Aug 17, 2017 at 17:54
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    $\begingroup$ I did not calculate a Fourier Transform, and i did not read the word FT in the question. My answer shows a regularization method (frequently used in physics). also your (correct) calculation leads to a result different from what the OP must obtain. The point is that given the ambiguity in the question the calculation that gives the result wanted by the OP, is preferable. Everything else is meaningless philosophy. $\endgroup$
    – Frank
    Aug 18, 2017 at 13:32
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    $\begingroup$ "The integral in question fails to exist." i never said otherwise.---- "The regularization you used produced a result that diverges for $\alpha=0$" so? it seems reasonable since when $\alpha=\epsilon=0$ it reduces to $\int_{R^+}1\,dx$ ---- "The answer you provided is incomplete its inverse Fourier transform does not recover the Heaviside function" the OP has never asked such Fourier transform. It seems pointless to respond to you once more. You like Fourier transforms, i get it! Peace. $\endgroup$
    – Frank
    Aug 19, 2017 at 0:47

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