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Topic: Incorrect symbolic improper integral
Replies: 11   Last Post: Oct 5, 2009 7:53 AM

 Messages: [ Previous | Next ]
 lshifr@gmail.com Posts: 531 Registered: 2/9/09
Re: Incorrect symbolic improper integral
Posted: Sep 30, 2009 7:33 AM

A follow - up to my previous post:

1. Of course, the simple form of the answer I gave is just

Pi*Exp[-Abs[a]]

2. When I was talking about pole contributions, I meant for two
exponential terms Exp[I*a*x] and Exp[-I*a*x] (expanding the cosine)
separately. The two terms always pick the opposite poles but give
the same contribution. The single term never gets the contribution from
both poles.

Regards,
Leonid

On Tue, Sep 29, 2009 at 4:38 AM, jwmerrill@gmail.com <jwmerrill@gmail.com>wrote:

> Below is a definite integral that Mathematica does incorrectly.
> Thought someone might like to know:
>
> In[62]:= Integrate[Cos[x]/(1 + x^2), {x, -\[Infinity], \[Infinity]}]
>
> Out[62]= \[Pi]/E
>
> What a pretty result--if it were true. The correct answer is \[Pi]*Cosh
> [1], which can be checked by adding a new parameter inside the
> argument of Cos and setting it to 1 at the end:
>
> In[61]:= Integrate[Cos[a x]/(1 + x^2), {x, -\[Infinity], \[Infinity]},
> Assumptions -> a \[Element] Reals]
>
> Out[61]= \[Pi] Cosh[a]
>
> Regards,
>
> Jason Merrill
>
>

Date Subject Author
9/30/09 Mark McClure
9/30/09 Andrzej Kozlowski
9/30/09 DrMajorBob
9/30/09 lshifr@gmail.com
9/30/09 jwmerrill@gmail.com
9/30/09 Daniel Lichtblau
10/3/09 Mariano
10/4/09 Daniel Lichtblau
10/5/09 Mariano
9/30/09 Bayard Webb
9/30/09 lshifr@gmail.com