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Bug persisting at least through version 11.3.0

The standard Gaussian integral

Integrate[Exp[-a x^2], {x,-Infinity,Infinity}]

returns

ConditionalExpression[Sqrt[Pi]/Sqrt[a], Re[a] > 0]

But if we manually substitute a = I, which apparently lies outside the integral's region of convergence, we get that

Integrate[Exp[-I x^2], {x,-Infinity,Infinity}]

returns

 (1 - I) Sqrt[Pi/2]

contradicting the answer to the more general problem, which implies that this integral does not converge. More generally,

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

returns

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

This Wikipedia article explains why the latter answers are correct. Why is the first result giving too small of a region of convergence?

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Wolfram Tech Support has confirmed that this is a bug in Mathematica.

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The expression is in fact regular for a=0 i.e. Re[a]=0. Assume

L > 0

and compute

Integrate[Exp[-a x^2], {x,-L,L}]

obtaining the function

(Sqrt[\[Pi]] Erf[Sqrt[a] L])/Sqrt[a]

which is regular in the point a=0, as Erf[x] ~ 2x/Sqrt[Pi] + O[x^3].

So as you correctly wrote for purely imaginary a, Re[a]=0 the integral converges.

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  • $\begingroup$ As $a \to 0$ the integral goes to $2L$, so it's regular for finite L, but the limit $L \to \infty$ required for the improper integral $\int_{-\infty}^\infty$ does not exist. The integral does not converge if $a = 0$. Anyway, it doesn't answer my question, which is why Mathematica is returning the wrong answer. $\endgroup$ – tparker Jul 4 '18 at 17:17

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