Cfrgtkky

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

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.

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.

$mathbf{UPDATE:}$

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.

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

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

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

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

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

Combining these:

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

This is consistent with what Mathematica (Version 11.2.0.0, Mac OS X) gives:

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

returning

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:

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.