
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 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

