Mathematica Integrate vs. NIntegrate in Mathematica

[nnnrocks]

I'm trying to integrate the function Exp[-I*t*x - t^2/2] from -infinity to infinity using NIntegrate in Mathematica; the value that I get is accurate when x is small, but as x gets larger, the output from NIntegrate does not match the value I get when I use Integrate -- it gets less and less accurate.
Does anyone know why this happens and what I can do to make sure NIntegrate is giving me accurate answers? (I'm ultimately going to apply NIntegrate to a much more complicated function and I'm using this test function to figure out how to deal with the oscillatory behavior)

 

[Hepth]

The exact result is $$ e^{-\frac{x^2}{2}} \left(\sqrt{2 \pi }\right) $$

Which for ##x == 10# is 4.83466*10^-22

So you need HIGH precision to recreate that with NIntegrate. You can do that, and set a precision goal by : (In the following the first example is NOT ENOUGH PRECISION, the second is enough )

With[{x = 10}, 
 NIntegrate[Exp[-I*t*x - t^2/2], {t, -\[Infinity], \[Infinity]}, 
  WorkingPrecision -> 20, PrecisionGoal -> 6]]

This is not enough, and gives

NIntegrate::ncvb: "\!\(\*
StyleBox[\"\\\"NIntegrate failed to converge to prescribed accuracy after \\\"\", \"MT\"]\)\!\(\* StyleBox[\"9\", \"MT\"]\)\!\(\* StyleBox[\"\\\" recursive bisections in \\\"\", \"MT\"]\)\!\(\* StyleBox[\"t\", \"MT\"]\)\!\(\* StyleBox[\"\\\" near \\\"\", \"MT\"]\)\!\(\* StyleBox[ RowBox[{\"{\", \"t\", \"}\"}], \"MT\"]\)\!\(\* StyleBox[\"\\\" = \\\"\", \"MT\"]\)\!\(\* StyleBox[ RowBox[{\"{\", \"1.255585675845388`\", \"}\"}], \"MT\"]\)\!\(\* StyleBox[\"\\\". NIntegrate obtained \\\"\", \"MT\"]\)\!\(\* StyleBox[ RowBox[{ RowBox[{\"-\", \"1.0581813203458523`*^-16\"1.0581813203458523`*}], \"+\",  RowBox[{\"3.469446951953614`*^-17\"3.469446951953614`*, \" \", \"I\"}]}], \"MT\"]\)\!\(\* StyleBox[\"\\\" and \\\"\", \"MT\"]\)\!\(\* StyleBox[\"6.409147557108418`*^-16\"6.409147557108418`*, \"MT\"]\)\!\(\* StyleBox[\"\"\", \"MT\"]\) for the integral and error estimates.

With[{x = 10}, 
 NIntegrate[Exp[-I*t*x - t^2/2], {t, -\[Infinity], \[Infinity]}, 
  WorkingPrecision -> 40, PrecisionGoal -> 6]]

This gives

 4.834658903596599769813455011618390931572*10^{-22} + 
 1.783512275559820938759410037056409076459*10^{-51} i

So youll have to increase your working precision depending on how high "x" is.

For x = 50, you'll have to have the working precision greater than 87 (90 works).