I have a simple gaussian integral: $\int^{\infty}_{-\infty}dx\:e^{i\alpha x^2}$.

If $\alpha \in \mathbb{R}$, then:

$$ \int^\infty_{-\infty} dx\; e^{i \, \alpha x^2} = \sqrt{\frac \pi {-i \alpha}} \qquad \qquad \alpha <0 $$

Now if $\alpha \in \mathbb{C}$ then we obtain the same answer but with different conditions:

$$ \int^\infty_{-\infty} dx\; e^{i \, \alpha x^2} = \sqrt{\frac \pi {-i \alpha}} \qquad \qquad Im(\alpha)>0 $$

These can be combined into a simple answer with an OR statement:

$$ \int^\infty_{-\infty} dx\; e^{i \, \alpha x^2} = \sqrt{\frac \pi {-i \alpha}} \qquad \qquad Im(\alpha)=0 \, \& \, Re(\alpha) <0 \quad|| \quad Im(\alpha)>0 $$

When I I ask Mathematica to solve this for me

Integrate[E^(I x^2 a), {x, -∞, ∞}]

Mathematica returns only one of these cases:

ConditionalExpression[Sqrt[π]/Sqrt[-I a], Im[a] > 0]

I have tried specifying $Im(\alpha)\geq 0$ in the Assumptions but Mathematica disregards this and gives the same result. It can of course find the answer for $\alpha \in \mathbb{R}$ but it cannot seem to give a fully general answer where both conditions are present.

  • $\begingroup$ Where did this integral come up? $\endgroup$ – mjw Mar 7 at 18:45

If $\alpha$ is complex, then yes, The imaginary part of alpha should be strictly greater than zero. If $\alpha$ is real, does it converge? If you complete the contour in the complex plane, with a semicircle, and replace $x$ by $z=x+ i y$, I do not see the integral converging along the semicircle.


Okay, I looked the integral up in Gradshteyn and Ryzhik, Table of Integrals, Series, and Products, 6$^\textrm{th}$ edition, p. 333. Looks like the integral for $\alpha$ real converges for $\alpha<0$ but with limits zero to infinity.

$\displaystyle \int_0^\infty e^{-i\lambda x^2} dx = \frac{1}{2} \sqrt{\frac{\pi}{\lambda}} e^{-i\pi/4}, \quad (\lambda>0)$.

We can infer from this (let $w=-x$ $\Rightarrow$ $dw = - dx$)

$\displaystyle \int_{-\infty}^\infty e^{-i\lambda x^2} dx = \sqrt{\frac{\pi}{\lambda}} e^{-i\pi/4}, \quad (\lambda>0)$.

Replacing $\lambda$ by $-\lambda$ we get the complex conjugate:

$\displaystyle \int_{-\infty}^\infty e^{i\lambda x^2} dx = \sqrt{\frac{\pi}{-\lambda}} e^{i\pi/4}, \quad (\lambda<0)$.

Combining these:

$\displaystyle \int_{-\infty}^\infty e^{-i\lambda x^2} dx = \sqrt{\frac{\pi}{\lambda}} e^{ - i\pi \,\textrm{sign }{(\lambda)} /4}, \quad (\lambda \ne 0, \lambda \in \Re)$.

This is consistent with what Mathematica (Version, Mac OS X) gives:

Assuming[Element[a,Reals],Integrate[Exp[I a x^2],{x,-Infinity,Infinity}]


Sqrt[Pi]/2 (1+ I Sign[a])/Sqrt[Abs[a]]

If anybody has an idea how to compute this integral with contour integration (or otherwise) from first principals, that would be interesting!

Also, we still haven't answered why Mathematica assumes that $\alpha$ is not real if $\alpha \in \mathbb{C}$!

$\mathbf{ADDITIONAL \,\, UPDATE:}$

This will take into account each case, and output the result or indicate if the integral does not converge:

q[a_] := Integrate[Exp[I a x^2], {x, -Infinity, Infinity}]; 
integral[a_] := If[Element[a, Reals], Assuming[Element[a, Reals], q[a]], q[a]];

Here are a few examples:

enter image description here

  • 1
    $\begingroup$ I don't think Mathematica gave the proper response $\alpha<0$. If it converges for $\alpha<0$ why not $\alpha>0$. and does it converge? $\endgroup$ – mjw Mar 7 at 15:38
  • $\begingroup$ I have assumed $x \in \mathbb{R}$ so convergence in that sense has not been accounted for. $\endgroup$ – OldTomMorris Mar 7 at 16:28
  • $\begingroup$ Well, I believe Gradshteyn and Ryzhik's listing. We can then make some substitutions to compute what happens for $\alpha>0$. Mathematica (ver. 11.2) gives an answer that is consistent. Obviously, the integral diverges for $\alpha=0$. Mathematica's result $\rightarrow \infty$ there. $\endgroup$ – mjw Mar 7 at 18:44
  • $\begingroup$ "Also, we still haven't answered why Mathematica assumes that 𝛼 is not real if 𝛼 ∈ ℂ!". This is exactly my problem! $\endgroup$ – OldTomMorris Mar 8 at 10:26
  • 1
    $\begingroup$ Agreed! I've updated my answer to output what we would have liked to have seen for cases when $\alpha$ is real and when $\alpha$ has a nonzero imaginary part. $\endgroup$ – mjw Mar 8 at 18:36

By default, Mathematica assumes all symbols are complex valued, so this is what you get if you don't specify. You can see all the variations by making assumptions:

Integrate[E^(I x^2 a), {x, -∞, ∞}, Assumptions -> #] 
          & /@ {a ∈ Complexes, a ∈ Reals, Im[a] == 0, 
                Im[a] < 0, Im[a] > 0, Re[a] == 0, Im[a] >= 0, a == 0, a < 0}

One of these doesn't converge and another is equal to infinity, but they do give the full range of possibilities.


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