Re: Simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg119266] Re: Simple integral
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Fri, 27 May 2011 06:14:31 -0400 (EDT)
On 5/26/11 at 7:17 AM, mariano.pierantozzi at gmail.com (Mariano
Pierantozzi) 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])))])
First, consider:
In[31]:= g =
Simplify[Integrate[1/(x^2 + b x + c), x] // TrigToExp, b^2 - 4
c > 0]
Out[31]= (Log[Sqrt[b^2 - 4*c] - b - 2*x] -
Log[Sqrt[b^2 - 4*c] + b + 2*x])/
Sqrt[b^2 - 4*c]
In[32]:= D[g, x] // Simplify
Out[32]= 1/(x*(b + x) + c)
So, the result obtained from Integrate is clearly the
anti-derivative of the integrand and is a correct solution to
the indefinite integral.
Now consider your hand solution
In[33]:= f =
1 (1/Sqrt[
b^2 - 4 c]) (Log[(2 x +
b - (Sqrt[b^2 - 4 c]))/((2 x + b + (Sqrt[b^2 - 4 c])))]);
D[f, x] // Simplify
Out[34]= 1/(x (b+x)+c)
which demonstrates both f and g are valid solutions to the
indefinite integral. But since,
In[35]:= FindInstance[g != f, {b, c, x}]
Out[35]= {{b -> 149/10 - (11*I)/2, c -> 59/10 - (62*I)/5,
x -> 49/2 - I/10}}
it is clear f and g are identical for all b, c and x.