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Math Help - An integral involving Hermite polynomials

  1. #1
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    An integral involving Hermite polynomials

    Hello,

    is there any way of calculating the following integral in closed form:

    \int_0^1{\mbox{H}_n(x) e^{-\frac{x^2}{2}}\mbox{d}x}?

    Thank you
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  2. #2
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    Quote Originally Posted by Heirot View Post
    Hello,

    is there any way of calculating the following integral in closed form:

    \int_0^1{\mbox{H}_n(x) e^{-\frac{x^2}{2}}\mbox{d}x}?

    Thank you
    Probaby not. But Rodrigue's formula might give you something useful: H_n(x) = (-1)^n e^{x^2/2} \frac{d^n}{dx^n} \left( e^{-x^2/2}\right).
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  3. #3
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    What if you plugged the Rodriguez formula into the integral? You'd get

    \displaystyle{\int_{0}^{1}H_{n}(x)\,e^{-\frac{x^{2}}{2}}\,dx=<br />
\int_{0}^{1} (-1)^n e^{\frac{x^{2}}{2}} \frac{d^n}{dx^n} \left( e^{-x^2/2}\right)\,e^{-\frac{x^{2}}{2}}\,dx=(-1)^{n}\int_{0}^{1}  \frac{d^n}{dx^n} \left( e^{-x^2/2}\right)\,dx.}

    Can you see where to go from here?
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  4. #4
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    Yes, the result is expressed in terms of (n-1)th derivative of an exponential. Is there an easy way to proced with the differentiation?
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  5. #5
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    What I would do is try to recognize the result in terms of a Hermite polynomial. That way, you wouldn't actually need to do the derivative. That would work if you are allowed to have your final answer in terms of Hermite polynomials. Do you follow me?
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  6. #6
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    The plot thickens... It seem that "my" Hermite polynomials are the physicist's version and have exp(-x^2) in the definition so the procedure wouldn't work that trivially. But nevertheless, I managed to evaluate the integral using recurrence relations. Thanks to the both of you!
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  7. #7
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    You're welcome. Have a good one!
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