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Re: Simple integral

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  Re: Simple integral Mariano Pierantozzi <mariano.pierantozzi@gmail.com> - 2011-05-26 11:47 +0000
    Re: Simple integral Stefan Salanski <wutchamacallit27@gmail.com> - 2011-05-27 10:13 +0000

#2738 — Re: Simple integral

I am grateful to all who answered me.
All of you have suggest to me to use the command
Simplify[% // TrigToExp, b^2 - 4 c > 0]
In this way i've this solution:
(Log[-b + Sqrt[b^2 - 4*c] - 2*x] - Log[b + Sqrt[b^2 - 4*c] + 2*x])/
 Sqrt[b^2 - 4*c].
But this is not the solution of the departure integral.
If "whit my hand" calculate the solution i've this result:
1(1/Sqrt[b^2 -
     4 c]) (Log[(2 x +
      b - (Sqrt[b^2 - 4 c]))/((2 x + b + (Sqrt[b^2 - 4 c])))])
And i thing it's not the same, because i've to do the limit x-->Infininity
and in this case the mathematica solution is, once again, in complex field.
Can you help me again?
Thank you again!
Mariano Pierantozzi
PhD Student
Energy Engineering

2011/5/23 Mariano Pierantozzi <mariano.pierantozzi@gmail.com>

> Hi,
> I've got some problem studing this simple integral:
> Integrate[1/(x^2 + b x + c), x].
> The Mathematica solution is:
> (2 ArcTan[(b + 2 x)/Sqrt[-b^2 + 4 c]])/Sqrt[-b^2 + 4 c]
>
> The problem is that my secon order polinomial have two real solutions, so
> my
> delta (-b^2 + 4 c) is greater than zero. In this case the denominator of
> the
> solution does not exist or exist in complex field, but my x is a volume...
> I try in this way
> Integrate[1/(x^2 + b x + c), x, Assumptions -> {-b^2 + 4 c < 0}], but
> nothing!
> I can't trasform my arcotangent in two log.
> In summary I would like to have such a solution:
> Integrate[1/(x^2 + 5 x + 6), x]
> Log[2 + x] - Log[3 + x]
> but in general form.
> Sorry for my english!
> Mariano Pierantozzi
> PhD Student
> Energy Engineering
>

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#2743

Your hand solution is simply the real part of Mathematica's solution.
(note that the real part of the Log of a negative number, is the log
of the absolute value.)
The imaginary part of the solution comes from integrating over the
singularities of the original function. Your function lends itself to
computation with a contour integral, and so it can be seen that when
one of the singularities is within your bounds of integration, there
is a pole in the complex plane and so the residue at that point must
be taken into account. See http://mathworld.wolfram.com/Contour.html .
Since these poles are the only source of the imaginary piece, a plot
of Im[ ] shows that its a step function.
Plot[{Re[fn[x] /. {b -> 3, c -> 2}], Im[fn[x] /. {b -> 3, c -> 2}]},
{x, -10, 10}]
So in practice, if you wanted to integrate from x1 to x2, where
neither singularity is in (x1,x2), Mathematica's answer gives you
exactly what your hand solution does (100% real). The only difference
is if you try to integrate with one or both of the singularities
inside (x1, x2) Mathematica also includes the imaginary part from the
contour integral (which is not an artifact or anything, its supposed
to be there). When you did it by hand, did you use complex analysis?
-Stefan S

On May 26, 7:47 am, Mariano Pierantozzi
<mariano.pieranto...@gmail.com> wrote:
> I am grateful to all who answered me.
> All of you have suggest to me to use the command
> Simplify[% // TrigToExp, b^2 - 4 c > 0]
> In this way i've this solution:
> (Log[-b + Sqrt[b^2 - 4*c] - 2*x] - Log[b + Sqrt[b^2 - 4*c] + 2*x])/
>  Sqrt[b^2 - 4*c].
> But this is not the solution of the departure integral.
> If "whit my hand" calculate the solution i've this result:
> 1(1/Sqrt[b^2 -
>      4 c]) (Log[(2 x +
>       b - (Sqrt[b^2 - 4 c]))/((2 x + b + (Sqrt[b^2 - 4 c])))])
> And i thing it's not the same, because i've to do the limit x-->Infininit=
y
> and in this case the mathematica solution is, once again, in complex fiel=
d.
> Can you help me again?
> Thank you again!
> Mariano Pierantozzi
> PhD Student
> Energy Engineering
>
> 2011/5/23 Mariano Pierantozzi <mariano.pieranto...@gmail.com>
>
>
>
>
>
>
>
> > Hi,
> > I've got some problem studing this simple integral:
> > Integrate[1/(x^2 + b x + c), x].
> > The Mathematica solution is:
> > (2 ArcTan[(b + 2 x)/Sqrt[-b^2 + 4 c]])/Sqrt[-b^2 + 4 c]
>
> > The problem is that my secon order polinomial have two real solutions, =
so
> > my
> > delta (-b^2 + 4 c) is greater than zero. In this case the denominator o=
f
> > the
> > solution does not exist or exist in complex field, but my x is a volume=
...
> > I try in this way
> > Integrate[1/(x^2 + b x + c), x, Assumptions -> {-b^2 + 4 c < 0}], but
> > nothing!
> > I can't trasform my arcotangent in two log.
> > In summary I would like to have such a solution:
> > Integrate[1/(x^2 + 5 x + 6), x]
> > Log[2 + x] - Log[3 + x]
> > but in general form.
> > Sorry for my english!
> > Mariano Pierantozzi
> > PhD Student
> > Energy Engineering

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