For $a \in \mathbb{R},a>0 $ the Gaussian integral is $$ \begin{equation} \int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac{\pi} {a}} . \hspace{1cm} (1) \end{equation} $$

What happens if we choose $a$ to be a complex number, let's say $-i$. Then, according to the formula above we would get (I am not sure if this is allowed ?) $$ \int_{-\infty}^{\infty} e^{ix^2} dx = \sqrt{i \pi} = \frac{(1+i)\sqrt{\pi}}{\sqrt{2}}.$$ But interestingly enough, Wolfram Alpha gives the same result.

Using Residue calculus we get

$$ \int_{-R}^{R} e^{ix^2} dx = - \int_{\gamma} e^{iz^2} dz, $$ where $\gamma$ is a semicircle in the upper half plane of radius R. With $t\to Re^{it}$ the rhs becomes:

$$ \int_{0}^{\pi} e^{iR^2 e^{2it}} iRe^{it} dt. $$ But how can that converge for $R \to \infty$? For example take $t=0$: We would have $ i Re^{iR^2}$, which for $R \to \infty$ goes to $\infty$ as $e^{ix^2}$ is bounded by $1$. Considering the above finite result, what is wrong here? Maybe we have to apply some distributional tools? Is it then just coincidence that formula $(1)$ worked?

Having stated this, can anyone help me out of this confusion?

  • $\begingroup$ Related. $\endgroup$ Jan 24, 2016 at 23:02
  • $\begingroup$ See the Fresnel Integral. - Mark $\endgroup$
    – Mark Viola
    Jan 24, 2016 at 23:05
  • 3
    $\begingroup$ Residue calculus doesn't magically work for any contour; the integral over the unwanted parts of the contour must tend to $0$. That doesn't happen here, at least not easily (almost always in these proofs, the integral of the modulus of the function over the rest of the contour goes to $0$, but that won't work here). On a related note, your integral does not converge absolutely, so it needs to be interpreted as a limit of finite integrals (as opposed to a Lebesgue integral); I think in this case, you can even let both limits tend independently to infinity. $\endgroup$ Jan 24, 2016 at 23:57

1 Answer 1


Yes you are fully right by setting that conjecture for $a=\pm i$ in fact there is more fancy way to that answer by hand computations.

Let me briefly explain the steps: it is more wise to think about Feymann's tricks by considering the the following( perturbed) Gaussian $$\color{blue}{e^{(-t+i)x^2} = e^{-tx^2}\cos(x^2)+ie^{-tx^2}\sin(x^2)}$$ then apply polar coordinates and let $t\to0$ aftermath to get the result. this is exactly what I did in my answer here (see all details below)

Proof: Indeed, we proceed as follow: Let, $$I=\int_0^\infty \cos(x^2) dx =\frac{1}{2}\int^\infty_{-\infty}\cos(x^2)dx~~~ and ~~~~ J=\int_0^\infty \sin(x^2) dx=\frac{1}{2}\int^\infty_{-\infty} \sin(x^2) \,dx $$ We set, $$I_t = \int_0^\infty e^{-tx^2}\cos(x^2) dx~~~~\text{and}~~~ J_t = \int_0^\infty e^{-tx^2}\sin(x^2) dx$$ $t\mapsto I_t$ and $t\mapsto J_t$ are clearly continuous due to the present of the integrand factor $e^{-tx^2}$.

Using Fubini we have,

\begin{split} I_t^2-J^2_t&=& \left(\int_0^\infty e^{-tx^2}\cos(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\cos(y^2) dy\right) \\&-& \left(\int_0^\infty e^{-tx^2}\sin(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\sin(y^2) dy\right) \\ &=& \int_0^\infty \int_0^\infty e^{-t(x^2+y^2)}\cos(x^2+y^2)dxdy\\ &=&\int_0^{\frac\pi2}\int_0^\infty re^{-tr^2}\cos r^2 drd\theta\\&=&\frac\pi4 Re\left( \int_0^\infty \left[\frac{1}{i-t}e^{(i-t)r^2}\right]' dr\right)\\ &=&\color{blue}{\frac\pi4\frac{t}{1+t^2}\to 0~~as ~~~t\to 0} \end{split}

However using similar technique in above prove one can easily arrives at the following $$\color{blue}{I_tJ_t = \frac\pi8\frac{1}{t^2+1}}$$ from which one get the following explicit value of $$\color{red}{I^2=J^2= IJ = \lim_{t\to 0}I_tJ_t =\frac{\pi}{8}}$$

having say that you will get, $$\int_{-\infty}^{\infty} e^{ix^2} dx = \sqrt{i \pi} = \frac{(1+i)\sqrt{\pi}}{\sqrt{2}}.$$


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