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Re: Incorrect symbolic improper integral
Posted:
Sep 30, 2009 7:41 AM


Isn't this integral easier to do analytically? The poles are at +/i. Write= cos(ax) as (Exp(iax)+Exp(iax))/2. First part you have to complete the con= tour around the upper half plane, second part round the lower (or vice vers= a if a is negative  all assumes a is real). All of which gives integral a= s pi Exp(Abs(a)), set a=1 and you get pi/e  as you say a pretty result! ________________________________________ From: mathgroupadm@smc.vnet.net [mathgroupadm@smc.vnet.net] On Behalf Of = Bayard Webb [bayard.webb@mac.com] Sent: Wednesday, September 30, 2009 9:59 AM Subject: Re: Incorrect symbolic improper integral
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 resultif 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



