
Re: are the commercial results continuous?
Posted:
Jul 29, 2012 12:52 PM


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*x2*x^2)/((1+x+x^2)^2*Sqrt[1x+x^2]),{x,1,0}] > > Integrate[(1x+3*x^2)/((1+x+x^2)^2*Sqrt[1x+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

