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Re: Discrepancy between Integrate and NIntegrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg83617] Re: Discrepancy between Integrate and NIntegrate
*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>
*Date*: Sun, 25 Nov 2007 04:31:40 -0500 (EST)
*References*: <fi8pjd$gti$1@smc.vnet.net>
chuck009 <dmilioto at comcast.com> wrote:
> Jesus David, . . . thanks. Nicely done. Here's the AoPS thread that
> this comes from:
>
> http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175132
>
> Maybe you could provide a comment there as well? Otherwise, I'll study
> your analysis and post one my own. Thanks again. :)
I don't think I'll be posting to the AoPS thread, but having looked at it,
I'll make another comment here.
The original integral considered there can be done by Mathematica directly
as a definite integral:
In[2]:= Integrate[Cos[x]/Cosh[x], {x, -Infinity, Infinity}]
Out[2]= Pi*Sech[Pi/2]
But it can also be done using Mathematica's indefinite integral and then
applying Newton-Leibniz:
In[6]:= Integrate[Cos[x]/Cosh[x], x]
Out[6]=
(1/2 + I/2)*E^((1 - I)*x)*(Hypergeometric2F1[1/2 - I/2, 1, 3/2 - I/2,
-E^(2*x)] - I*E^(2*I*x)*Hypergeometric2F1[1/2 + I/2, 1, 3/2 + I/2,
-E^(2*x)])
In[7]:= indef = FullSimplify[%, Element[x, Reals]]
Out[7]=
((-(1/2))*I*((-1)^I*Beta[-E^(2*x), 1/2 - I/2, 0] + Beta[-E^(2*x), 1/2 +
I/2, 0]))/(-1)^(I/2)
In[8]:= indef /. x -> -Infinity
Out[8]= 0
In[9]:= Limit[indef, x -> Infinity]
Out[9]=
(1/2 + I/2)*(Gamma[1/2 + I/2]*Gamma[3/2 - I/2] - I*Gamma[1/2 -
I/2]*Gamma[3/2 + I/2])
In[10]:= FullSimplify[%]
Out[10]= Pi*Sech[Pi/2]
The above worked correctly because Mathematica's indefinite integral gave
an antiderivative which was valid on the whole real line.
If you have any further questions, let me know.
David
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