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Re: Integrate (undocumented feature)

  • To: mathgroup at
  • Subject: [mg20787] Re: Integrate (undocumented feature)
  • From: "Bill Bertram" <wkb at>
  • Date: Thu, 11 Nov 1999 00:22:55 -0500
  • Organization: Australian Nuclear Science and Technology Organisation
  • References: <80asap$>
  • Sender: owner-wri-mathgroup at

Ersek, Ted R wrote in message <80asap$it2 at>...
>Earlier I wrote about the following results using Version 4.
>The documentation for NIntegrate says:
>"NIntegrate[f, {x,x0,x1, ... ,xk}] tests for singularities at each of the
>intermediate points xi. If there are no singularities, the result is
>equivalent to an integral from x0 to xk. You can use complex numbers xi to
>specify an integration contour in the complex plane."
>Although the documentation doesn't say so it seems this applies to
>as well.  When I wrote the previous email I was thinking of the line above,
>but didn't remember that I read this in the documentation for NIntegrate
>Below I give convincing evidence that this works with Integrate. Here I
>integrate along a closed contour in the complex plane.  Notice I get the
>same answer when I apply a theorem related to Residues.
>    {z,-1,1,1+3I,-1+3I,-1}]
>(2 Pi I)Residue[1/(z^2+4),{z,2I}]

>I think the documentation for Integrate should be changed to mention this

Or perhaps it should be avoided altogether!
Consider the following,

Integrate[1/(x - 2I), {x, -1, 1, 1 + 3I, -1 + 3I, -1}] // FullSimplify

this gives  the result  I Pi (wrong by a factor 2)

And upon changing the contour  (still closed and around the singularity)

    1/(x - 2I), {x, -1 - I, 1 - I, 1 + 3I, -1 + 3I, -1 - I}] // FullSimplify

the result is 0.

Not the sort of results to inspire confidence in Mathematica's integration

(NIntegrate does give the correct result for both contours however)



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