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Math Help - Integral - no primitive?

  1. #1
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    Integral - no primitive?

    Hello...

    I'm struggling with this statistic integral that I believe does NOT have a primitive function.
    Or at least I cannot find it without the numerical root of pi (which I need to avoid).

    Can someone please double check and confirm or correct my statement?
    If I'm wrong, please show me how to get to the primitive. THANKS!

    <br />
\int_0^\infty 2 \lambda x\ e^{-\lambda(x-\frac{t}{2\lambda})^2} dx\<br />
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  2. #2
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    Quote Originally Posted by paolopiace View Post
    Hello...

    I'm struggling with this statistic integral that I believe does NOT have a primitive function.
    Or at least I cannot find it without the numerical root of pi (which I need to avoid).

    Can someone please double check and confirm or correct my statement?
    If I'm wrong, please show me how to get to the primitive. THANKS!

    <br />
\int_0^\infty 2 \lambda x\ e^{-\lambda(x-\frac{t}{2\lambda})^2} dx\<br />
    Make the substitution u = x - \frac{t}{\lambda} and split the integral up into two bits.

    Hopefully it's clear that one bit is simple .....

    The other bit requires the integral of a simple Gaussian from -\frac{t}{\lambda} to infinity. Not so simple .... it is well known in fact that no exact answer can be calculated (if it was from 0 to infinity it could be done easily).
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    Thanks but...

    ... it seems that your reply confirms my statement: there is no primitive. Correct?

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

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

    Then I changed variables:

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

    That gave a numeric value of this integral equal to

    <br />
\frac{t}{2}\sqrt\frac{\pi}{\lambda}<br />

    But I don't want a numeric value.
    I need a primitive that, for a gaussian, does not exist...

    Comments are super welcome.
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  4. #4
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    Quote Originally Posted by paolopiace View Post
    ... 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:

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

    Then I changed variables:

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

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

    That gave a numeric value of this integral equal to

    <br />
\frac{t}{2}\sqrt\frac{\pi}{\lambda}<br />

    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 z = -\frac{t}{2 \sqrt{\lambda}} and 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.
    As for that why you might find this thread enlightning.
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    Mr F, thanks...

    I see your point on the changes of variable.
    I did correct my calculus (changing the starting point of the integral) and found the exact same results.

    I'm changing approach now.
    Thanks again.
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