hello,i`m kinda stuck with this integral,

i wrote my approach,but it didnt get me too far.

thanks!

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- Jan 26th 2009, 04:18 PMbecker89evaluate integral
hello,i`m kinda stuck with this integral,

i wrote my approach,but it didnt get me too far.

thanks! - Jan 26th 2009, 04:39 PMMush
Ermmm...

- Jan 26th 2009, 04:41 PMbecker89
i suppose there is no analytic solution for that thing then?

how much reliable really is that integrator? - Jan 26th 2009, 04:43 PMMush
It's Wolfram - possibly the best 'automatic' integrator in the world. I just tested it in Matlab, and the answer given there is equally as dubious. In fact Matlab says "warning, explicit integral could not be found." So I'd definitely say, yes, there is no analytic solution.

- Jan 26th 2009, 04:49 PMMush
Where did you find such a question anyway?

- Jan 26th 2009, 04:49 PMbecker89
well i have another integral,i think putting it here is better than opening a new thread.

this integral is something unique,according to the person who uploaded it,

there is a solution using elementary function,

even though the fact that if you plug it in to different integrators,

it would show a messy result (like the integral i asked about earlier).

the factoring i did is a suggestion i got from the same person

many thanks! - Jan 26th 2009, 04:50 PMbecker89
- Jan 28th 2009, 02:14 AMsimplependulum
The four complex roots of t^4 - 2t^3 + t^2 + 6t + 3 are

[sqrt(6) + 1 ]/2 +/- [sqrt(3) + sqrt(2) ]/2 x sqrt(-1) and

[1 - sqrt(6)]/2 +/- [ sqrt(3) - sqrt(2) ] /2 x sqrt(-1)

They are two conjugate pairs so that

t^4 - 2t^3 + t^2 + 6t + 3 can be factorized into two quadartic polynomials .

.........

the result i found is

(k1) ln| t^2 - [sqrt(6) + 1]t + 3 + sqrt(6) | +

(k2) ln| t^2 - [1 - sqrt(6)]t + 3 - sqrt(6) | +

(k3) arctan{ [sqrt(3) - sqrt(2)][2t - sqrt(6) - 1 ] } +

(k4) arctan{ [sqrt(3) + sqrt(2)][2t + sqrt(6) - 1] } + C

where k1,k2,k3,k4 are constant and t = x + sqrt ( x^2 + x + 1)

I'm not sure ...

k1 = -0.237

k2 = -0.0160

k3 = -0.823

k4 = -0.0555