integrate(sin(t),t,minf,inf) = 0?

• Subject: integrate(sin(t),t,minf,inf) = 0?
• From: Raymond Toy
• Date: Tue, 07 Sep 2010 13:07:47 -0400

 The current CVS version of maxima says

integrate(sin(t),t,minf,inf) -> 0

I don't think that's right.  0 would be the Cauchy principal value, I
think, but not the value of the improper integral.  Maxima doesn't
normally use the Cauchy principal value without at least telling the
user that it is the principal value.

Ray

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