Ygthkb

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.

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, -[Infinity], [Infinity]}, Assumptions -> #] 

          & /@ {a [Element] Complexes, a [Element] 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.

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.

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

We can infer from this

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

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

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

Combining these:

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

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 $alpha in mathbb{C}$!