exponential integral

**abotaha** · July 18th 2010, 02:37 AM

Dear all,
I have exponential integral with polynomial. I tried to solve it but I could not

.
the integration is :

$\int_{-\infty}^{\infty} (ax^2+bx+c) e^{ax^2+bx+c} dx$

Can any one help me please.

Thanks in advance,

**Prove It** · July 18th 2010, 03:09 AM

Wolfram couldn't give me an answer...

Are you sure it's not

$\int_{-\infty}^{\infty}(2ax + b)e^{ax^2 + bx + c}\,dx$

because that can be solved with a substitution...

Actually on second thought, I believe that integral I mentioned is divergent... Hmmm...

**abotaha** · July 18th 2010, 03:39 AM

hi,
thank you for the replay.
yes the integral that i wrote is correct. and if it is like what you wrote then you are writ that it can be solved with substitution.

any other suggestion please.

**mr fantastic** · July 18th 2010, 03:41 AM

Originally Posted by abotaha

Dear all,
I have exponential integral with polynomial. I tried to solve it but I could not

.
the integration is :

$\int_{-\infty}^{\infty} (ax^2+bx+c) e^{ax^2+bx+c} dx$

Can any one help me please.

Thanks in advance,

Where has this integral come from?

**abotaha** · July 18th 2010, 04:10 AM

ok, there are mistakes in the integral equation.
the correct one is :

$\int_{-\infty}^{\infty} (-ax^2+bx+c) e^{-dx^2+bx+f} dx$

sorry for this mis-writing.

please help me to solve this integration.

**mr fantastic** · July 18th 2010, 04:12 AM

Originally Posted by abotaha

ok, there are mistakes in the integral equation.
the correct one is :

$\int_{-\infty}^{\infty} (-ax^2+bx+c) e^{-dx^2+bx+f} dx$

sorry for this mis-writing.

please help me to solve this integration.

You still haven't answered my question:

Originally Posted by Mr Fantastic

Where has this integral come from?

**abotaha** · July 18th 2010, 05:44 AM

The integral comes from multiplication of two functions:

$f_1=\ln (\frac {1}{4\pi^2})- \frac {(y-\mu_y-\beta_{yx}(x-\mu_x) -\beta_{yz}( z-\mu_z)-\beta_{yw}(w-\mu_w))^2}{2 \sigma_y^2} - \frac {( x-\mu_x )^2}{2 \sigma_x^2} - \frac {(z-\mu_z)^2}{2 \sigma_z^2} - \frac {(w-\mu_w)^2}{2 \sigma_w^2}- \ln(\sigma_y) - \ln(\sigma_x) - \ln(\sigma_z)- \ln(\sigma_w)$
and
$f_2=\frac{1}{2 \sqrt{\pi} \sigma_y\sigma_m}\left(\sqrt{2}\sqrt{\sigma_y^2+\b eta{yx}^2\sigma_m}\\<br /> e^T\right)$

where
$T=R+\left(\frac{(y-\mu_y+\beta_{yx}\mu_m-\beta_{yz}(z-\mu_z)-\beta_{yw}(w-\mu_w))^2}{2\sigma_y^2}+\frac{\mu_m^2}{2\sigma_m^2 }- <br /> \frac{\frac{(y-\mu_y+\beta_{yx}\mu_m-\beta_{yz}(z-\mu_z)-\beta_{yw}(w-\mu_w))\beta_{yx}}{\sigma_y^2}+\frac{\mu_m}{2\sigm a_m^2}}<br /> {2(\frac{\beta_{yx}^2\sigma_m^2+\sigma_y^2}{\sigma _y^2\sigma_m^2)}}\right)$
and
$R=\left(\frac {(y-\mu_y-\beta_{yx}(x-\mu_m) -\beta_{yz}( z-\mu_z)-\beta_{yw}(w-\mu_w))^2}{2 \sigma_y^2} - \frac {( x-\mu_m )^2}{2 \sigma_m^2}\right)$

then I simplified the multiplication result as it written above:
$\int_{-\infty}^{\infty}f_1f_2 dx$

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