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Re: Incorrect symbolic improper integral
Posted:
Sep 30, 2009 7:41 AM
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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: mathgroup-adm@smc.vnet.net [mathgroup-adm@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 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
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