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Re: Integration Problem

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  • Subject: [mg126368] Re: Integration Problem
  • From: Oliver Jennrich <oliver.jennrich at>
  • Date: Fri, 4 May 2012 06:25:35 -0400 (EDT)
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  • References: <jntg1g$g$>

Michael Musheghian <michael.musheghian at> writes:

> Greetings!
> I found that evaluation of this 2 integrals yield a bit different result. What could be the reason?
> Integrate[E^(-1/10 ((1 + r2z)^2)), {r2z, -Infinity, Infinity}]
> Integrate[E^(-0.1 ((1 + r2z)^2)), {r2z, -Infinity, Infinity}]

Numerics. The first integral evaluates symbolically, the second one
Mathematica 8 yields Sqrt[10 \[Pi]] for the former and 
5.60499 - 2.03152*10^-16 I for the latter. 

You can avoid the very small imaginary part by either calculating the
equivalent integral 

Integrate[E^(-0.1 ((r2z)^2)), {r2z, -Infinity, Infinity}] 

(i.e. performing a shift in the integration variable) or by having Mathematica calculate the integral fully symbolically:

Integrate[E^(-a ((r2z)^2)), {r2z, -Infinity, Infinity}, 
 Assumptions -> {a > 0}]

which yields not surprisingly Sqrt[\[Pi]]/Sqrt[a]

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