Originally Posted by

**paolopiace** ... it seems that your reply confirms my statement: there is no primitive. Correct? Mr F says: Correct.

Actually, I did go on with some further simplificatios nailing it down to:

$\displaystyle

t\int_0^\infty e^{-\lambda(x-\frac{t}{2\lambda})^2} dx\

$

Then I changed variables:

$\displaystyle

\sqrt\lambda(x-\frac{t}{2\lambda})=z;\ dx = \frac{1}{\lambda} dz

$

Mr F asks: Did you remember to change the integral terminals as well ......?

That gave a numeric value of this integral equal to

$\displaystyle

\frac{t}{2}\sqrt\frac{\pi}{\lambda}

$

Mr F answers his own question: From the fact that you got a numeric answer, I see you forgot. You used 0 to infinity, didn't you. The new integral terminals are $\displaystyle z = -\frac{t}{2 \sqrt{\lambda}}$ and $\displaystyle z = + \infty$. You cannot get an exact numeric value with those terminals. It just can't be done.

But I don't want a numeric value.

I need a primitive that, for a gaussian, does not exist... Mr F says: **It is well known that the integral of a Gaussian function has no elementary primitive.** Seeking one is futile. You have a better chance of flapping your arms and flying.

Comments are super welcome. Mr F says: I really don't know what sort of comment, super or otherwise, you're seeking. What you want to do is mathematically impossible. You know that. The only thing that can be added is *why*.