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

 Messages: [ Previous | Next ]
 Bayard Webb Posts: 5 Registered: 9/21/09
Re: Incorrect symbolic improper integral
Posted: Sep 30, 2009 7:33 AM

I think you need to add a as a coefficient of x everywhere, including
the squared term.

In[6]:= Assuming[a \[Element] Reals,
Integrate[Cos[a x]/(1 + (a x)^2), {x, -\[Infinity], \[Infinity]}]]

Out[6]= \[Pi]/(E Abs[a])

Setting a = 1 yields Mathematica's previous result.

Bayard

On Sep 29, 2009, at 4:38 AM, 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