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Re: are the commercial results continuous?
Posted:
Jul 29, 2012 12:52 PM
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Am 16.07.2012 22:50, schrieb clicliclic@freenet.de:
>
> clicliclic@freenet.de schrieb:
>>
>> [...]
>>
>> Only Mathematica results for the last two examples are missing now (I
>> tried Wolfram Alpha, and again it times out)!
>>
>
> Here they are again:
>
> Integrate[(2+3*x-2*x^2)/((1+x+x^2)^2*Sqrt[1-x+x^2]),{x,-1,0}]
>
> Integrate[(1-x+3*x^2)/((1+x+x^2)^2*Sqrt[1-x+x^2]),{x,-1,0}]
>
> and the correct results would be 0.06143194687 and 2.602719582.
>
> Martin.
>
I know it has been a while ago. But this is interesting:
Mathematica is such an extraordinary excellent software - the integrals
have been far too simple for it ;)
Introducing a pertubation in the denominator:
In[1]:= Integrate[
(2 + 3*x - 2*x^2)/((1 + a*x + x^2)^2*Sqrt[1 - x + x^2]),
{x, -1, 0},
Assumptions -> -2 < a < 2
] /. a -> 1.`20.
Out[1]= 0.06143194687872670 + 0.*10^-18 I
In[2]:= Integrate[
(1 - x + 3*x^2)/((1 + a*x + x^2)^2*Sqrt[1 - x + x^2]),
{x, -1, 0},
Assumptions -> -2 < a < 2
] /. a -> 1.`20.
Out[2]= 2.60271958298175737 + 0.*10^-18 I
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