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How does one evaluate $$\int_{-\infty}^{\infty} e^{ix^2} dx$$

I know the trick how to evaluate $\int_{-\infty}^{\infty} e^{-x^2}dx$ but trying to apply it here I get a limit which does not converge:

$I = \int_{-\infty}^{\infty} e^{ix^2}dx = \int_{-\infty}^{\infty} e^{iy^2}dy \\\implies I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{i(x^2+y^2)}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} re^{ir^2} = -\pi i (e^{i\infty}-e^0) $
and $e^{i\infty} $ is not defined.

Are there any other methods? I am not interested in the result (WolframAlpha can do this for me), but rather the method.

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    $\begingroup$ $|e^{i x^2}| = 1$ for all $x$, so it can't converge in the usual sense. $\endgroup$
    – copper.hat
    Nov 10, 2015 at 19:34
  • $\begingroup$ Hint: integrate around a wedge with angle $\pi/4$ in the first quadrant in the comlplex plane... $\endgroup$
    – tired
    Nov 10, 2015 at 19:35
  • $\begingroup$ the answer should be $$(1+i) \sqrt{\frac{\pi }{2}}$$ $\endgroup$ Nov 10, 2015 at 19:35
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    $\begingroup$ This is known as the Fresnel integral. $\endgroup$
    – copper.hat
    Nov 10, 2015 at 19:39
  • $\begingroup$ Yes, I am trying to solve the fresnel-integrals that way $\endgroup$
    – Christian
    Nov 10, 2015 at 20:56

2 Answers 2

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All right, let's do some complex analysis! Let's integrate $e^{iz^2}$ over the closed contour defined in three pieces (the arrows indicating the direction of contour integration) $$ \begin{cases} \Gamma_1: & |z|:0\rightarrow R, & \theta=0 \\ \Gamma_2: & |z| = R, & \theta: 0\rightarrow \pi/4\\ \Gamma_3: & |z|:R\rightarrow0, & \theta=\pi/4 \end{cases} $$ which we will eventually want to take the limit $R\rightarrow\infty$.

It can be seen that $$ \int_0^\infty e^{ix^2}dx=\lim\limits_{R\rightarrow\infty}\int_{\Gamma_1}e^{iz^2}dz $$ and since $e^{ix^2}$ is an even function $$ \int_{-\infty}^{\infty}e^{ix^2}dx=2\int_0^\infty e^{ix^2}dx $$ so we're heading in the right direction.

Now Cauchy's Theorem states that $$ \oint_D f(z)dz =0 $$ for $f(z)$ analytic in $D$. Our function, $e^{iz^2}$, has no singularities and is defined on the entire complex plane, so it is considered an entire function, and Cauchy's Theorem holds for our closed contour: $$ \int_0^R e^{ix^2}dx+\int_{\Gamma_2}e^{iz^2}dz+\int_{\Gamma_3}e^{iz^2}dz=0 $$

For our second integral above, we show that it vanishes as $R\rightarrow\infty$ using the ML test given by $$ \left|\int_\Gamma f(z)dz\right|\leq ML $$ where $M$ is a finite upper bound of $f(z)$ and $L$ is the length of the contour $\Gamma$. Of course, we need to assume that $f(z)$ is bounded and analytic on $\Gamma$ for this. In order to apply the ML test, we substitute into our integrand $z=re^{i\theta}$ so that $$ z^2 = r^2e^{2i\theta} = r^2cos(2\theta)+ir^2sin(2\theta) $$ $$ |e^{iz^2}|=|e^{ir^2cos(2\theta)-r^2sin(2\theta)}|\leq e^{-R^2}=M $$ because $r=R$ on this contour and $sin(2\theta)\leq1$. While, $$ L=\frac{\pi R}{4} $$ since we are looking at $1/8$th of the perimeter of the circle with radius $R$. By the ML test $$ \left|\int_{\Gamma_2} e^{iz^2}dz\right| \leq e^{-R^2}\frac{\pi R}{4} $$ which goes to $0$ as $R\rightarrow\infty$.

Now we want to deal with the 3rd contour integral $\Gamma_3$. Fortunately, the contour we picked allows us to easily parameterize this integral, as $y=x$. We will also need $z^2=(x+iy)^2=x^2-y^2+2ixy$. Recalling that $dz=dx+idy$ the integral becomes $$ \int_{\Gamma_3} e^{i(x^2-y^2)-2xy}(dx+idy) =\int_{R}^{0} e^{-2x^2}dx+i\int_{R}^{0} e^{-2y^2}dy \rightarrow-\sqrt{\frac{\pi}{8}}(1+i)\ \text{as}\ R\rightarrow0 $$ from our real Gaussian integral identities.

Taking $R\rightarrow\infty$, our results for the contour integrals in our Cauchy's Theorem equation imply that $$ \int_0^\infty e^{ix^2}dx = \sqrt{\frac{\pi}{8}}(1+i) $$

The integral from $-\infty$ to $\infty$ is just twice this. So boom.

If you want, you can rewrite $e^{ix^2}=cos(x^2)+isin(x^2)$ and equate the real and imaginary parts in the last equation and you will get the limiting values of the Fresnel Integrals.

Boom.

Also, since $$ (1+i)=\sqrt{2}e^{i\pi/4}=\sqrt{2e^{i\pi/2}}=\sqrt{2i} $$ we have $$ \int_0^\infty e^{ix^2}dx = \sqrt{\frac{i\pi}{4}} = \frac{1}{2}\sqrt{-\frac{\pi}{i}} $$

which exactly matches the well-known Gaussian integral identity $$ \int_0^\infty e^{-\alpha x^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}} $$ with $\alpha=-i$. Boom. Thus, this suggests that this identity can work for imaginary $\alpha$, and possibly certain complex $\alpha$ with the right combination of real and imaginary parts as well as choice of contours that do not make our integrals blow up.

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  • $\begingroup$ Hi, to avoid the technicalities in those case, then say (using integration by parts) that $\int_0^\infty e^{-i x^2}dx = \lim_{\epsilon \to 0^+}\int_0^\infty e^{-(i+\epsilon) x^2}dx$, and $\int_{-\infty}^\infty e^{-a x^2}dx = a^{-1/2} \int_{-\infty}^\infty e^{- x^2}dx$ for $a > 0$ and for $\Re(a) > 0$ by analytic continuation. $\endgroup$
    – reuns
    Oct 21, 2017 at 5:06
  • $\begingroup$ Nice results $\int_{-\infty}^\infty e^{ix^2}dx = \sqrt{\pi i}$. $\endgroup$
    – MathArt
    Aug 21, 2020 at 19:55
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    $\begingroup$ The ML argument is wrong. You can only say $\sin(2\theta) \ge 0 \implies M \le e^0 = 1$. Funnily enough, the same mistake was in the textbook I am using which is why I came to look for correct solution on MSE. I think I came up with a workaround though. First make a substitution $x^2\ = t$. Then you will have $\sqrt{t}$ in the denominator and you can show that the integral on the arch tends to zero. Then you come back to original coordinates. $\endgroup$ Oct 30, 2020 at 11:02
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    $\begingroup$ @RodionZaytsev is correct, the ML argument is wrong. We should use $sin(\theta)\geq \frac{2\theta}{\pi}$ for $\theta\in[0,\frac{\pi}{2}]$ and obtain $R\int_0^{\pi/2} e^{-R^2\theta}d\theta$ which goes to zero for $R\to\infty$. $\endgroup$
    – Kolja
    Mar 27, 2021 at 22:34
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    $\begingroup$ Boom, Boom, Boom. Strongly attacked that integral. $\endgroup$
    – Leon
    Aug 24, 2021 at 21:10
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Uh so I was presented this integral a while ago and also used a contour to solve it before I realized you can just use $e^{i\theta}=\operatorname{cis}(\theta)$. I thought I'd share this method ig.

Using the aforementioned property, the integral becomes \begin{align*} I=\int_{\mathbb{R}}e^{ix^2}\text{ d}x&=\int_{\mathbb{R}}\cos(x^2)+i\sin(x^2)\text{ d}x\\ &=\int_{\mathbb{R}}\cos(x^2)\text{ d}x+i\int_{\mathbb{R}}\sin(x^2)\text{ d}x\\ \end{align*} We can use our fresnel integral properties to find these integrals, and we see both of them result in $$\int_{\mathbb{R}}\cos(x^2)\text{ d}x=\int_{\mathbb{R}}\sin(x^2)\text{ d}x=\sqrt{\frac{\pi}{2}}$$ so our answer is just $$I=\sqrt{\frac{\pi}{2}}+i\cdot\sqrt{\frac{\pi}{2}}=\boxed{(1+i)\sqrt{\frac{\pi}{2}}}$$

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    $\begingroup$ This begs the question.. $\endgroup$ May 1 at 18:50
  • $\begingroup$ @CameronWilliams ??? what do you mean $\endgroup$
    – Max0815
    May 1 at 23:44
  • $\begingroup$ Your solution is circular. Many times those equalities are proved through the question OP posed. $\endgroup$ May 2 at 3:26
  • $\begingroup$ Hm fair ig. whatever $\endgroup$
    – Max0815
    May 2 at 4:12

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