What are some methods for solving the integral of (4x^2-1)e^(-2x^2)?

[birdhen]

I was wondering if anyone could point me in the right direction with this integral.

((4x^2)-1)e^(-2x^2),

I have tried substitution with trig functions, hyperbolic functions, seperating the first part into partial fractions and numerous other methods to no avail. Does anyone who knows have any hints?

Cheers

 

[yyat]

Try solving this with the http://en.wikipedia.org/wiki/Error_function" . This will give a solution in terms of erf(x), a solution in terms of more elementary functions is probably not possible.
I also suspect that the definite integral over the real line is 0.

 

[birdhen]

It is OK I have done it now, I split it into two parts 4x^2(e-2x^2) and -e(-2x^2) and integrated the first part by parts using u'=4xe(-2x^2) and v=x, and then the integral part of the solution canceled with the second part ie the sol'n was ;

[-xe(-2x^2)]-int(-e(-2x^2)-int(-e(-2x^2)),

leaving [-xe(-2x^2)] with appropriate limits,

Thanks!

 

[HallsofIvy]

It is OK I have done it now, I split it into two parts 4x^2(e-2x^2) and -e(-2x^2) and integrated the first part by parts using u'=4xe(-2x^2) and v=x, and then the integral part of the solution canceled with the second part ie the sol'n was ;

[-xe(-2x^2)]-int(-e(-2x^2)-int(-e(-2x^2)),

No, the last two terms do NOT cancel, they add. This is
$$-xe^{-2x^2}- 2\int e^{-2x^2} dx$$.

leaving [-xe(-2x^2)] with appropriate limits,

Thanks!

 

[birdhen]

I think wrote it incorrectly, it would be;

(-xe^(-2x^2)) -int(-e^(-2x^2)) -int(e^(-2x^2)),

which then cancels.