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I'm trying to calculate the following integral

$$ \int_0^\infty dx \, e^{ix}$$

with contour integration. By choosing a quarter circle contour $\gamma$, by the residue theorem we get

\begin{align} 0 =\oint_\gamma dz \, e^{iz} =& \int_0^\infty dx\, e^{ix} + \int_{\mathrm{Arc}} dz \, e^{iz} -i\int_0^\infty dy \, e^{-y} \\ =& \int_0^\infty dx\, e^{ix} + \int_{\mathrm{Arc}} dz \, e^{iz} -i \end{align}

therefore, if the integral over the arc is zero, the original integral becomes $i$.

Can Jordan's lemma, of a variation thereof, be applied in this case?

Edit: the parametrization would be $z=R e^{i\theta}$ with $\theta \in [0, \frac{\pi}{2}]$.

Aside: it seems to me that the original integral can be interpreted as a distribution as the Fourier transform of the theta function at $k=1$, which would give $i$ as well.

user35319
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  • If this is actually meant as a standard integral, then it is not convergent, since the integrand has constant norm 1. – Richard Jensen Jan 15 '21 at 09:51
  • What do you mean by standard integral? $e^{ix^2}$ also has norm 1, but it converges. – user35319 Jan 15 '21 at 09:56
  • I mean the Riemann or Lesbegue intergral. The integral of $e^{ix^2}$ from 0 to infinity also doesn't converge, since you can find arbitrarily large numbers $N$ such that

    $\int_0^N e^{ix^2}dx = 0$ or $\left|\int_0^N e^{ix^2}dx\right| = \left|\int_0^{\pi}e^{ix}\right|>0$

     – Richard Jensen Jan 15 '21 at 10:03
  • Also, you could just argue by the tail lemma for definate integrals (the sequence $\int_N^{\infty}f(x)dx$ goes to 0 for $N$ going to infinity if the integral is convergent). – Richard Jensen Jan 15 '21 at 10:08
  • @RichardJensen Such a sequence does not exist, since both the real and imaginary parts are strictly positive, as they are the Fresnel integrals https://en.wikipedia.org/wiki/Fresnel_integral. There are many answers which provide a computation of the integral of $e^{ix^2}$ already https://math.stackexchange.com/questions/1952056/computing-the-integral-int-expix2-dx, https://math.stackexchange.com/questions/1522926/integral-of-eix2 – user35319 Jan 16 '21 at 09:18

  • No, $e^{i\sqrt{\pi}^2}= - 1$, so has negative real part.

  • The wikipedia article's calculation for the infinity limit is nonsense, since it evaluates cosine at infinity, which is undefined.

  • I haven't read the entire argument from the SE post, but the top comment to the answer states the same that I do.

  • Check @Nico Terry's answer, there you can see exactly what goes wrong.

  • I'm guessing this is a problem from physics or enginering? The reason I'm thinking this is that in those disciplines, people are loose with math, and use "wrong" resulats if they just work.

    • – Richard Jensen Jan 16 '21 at 14:00
  • @RichardJensen what I mean was that $F(x)=\int_0^x dt , e^{it^2}$ has positive real and imaginary parts for all $x$. Are you impliying that the plots $S(x)$ and $C(x)$ on wikipedia are wrong? The integral of $e^{ix}$ is indeed undefined though, because as you say cosine and sine have no asymptotic limit – user35319 Jan 16 '21 at 15:15
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    Okay, I see. Well, it has non-negativ real and imaginary part, since it evaluates to 0 at for example 0 and $\sqrt{2\pi}$. And it turns out that I read the article on wikipedia wrong. Those Fresner integrals are indeed convergent (there is a misprint on the wikipedia article which confused me where they use $e^{t^2}$ instead of the correct $e^{-t^2}$, but I digress), so apologies for that, you were indeed right. I remembered the theorem wrong, the integrand does not have to go to zero for the integral to be convergent! – Richard Jensen Jan 16 '21 at 16:01
  • @RichardJensen by the way I was indeed asking from a physics background! I guess physicists don't care about convergence as always – user35319 Jan 16 '21 at 18:41
  • It is divergent but concerning distribution theory see this thread or this one. – Raymond Manzoni Jan 20 '21 at 11:12