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Re: Discrepancy between Integrate and NIntegrate

chuck009 <dmilioto at> wrote:
> Jesus David, . . . thanks.  Nicely done.  Here's the AoPS thread that
> this comes from:
> 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]
(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,

In[7]:= indef = FullSimplify[%, Element[x, Reals]]
((-(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]
(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.


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