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Math Help - Integrate second derivative squared.

  1. #1
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    Show that

    \int_{- \infty}^{+ \infty} (f''(x))^2 \, dx

    where

     <br />
f(x)=\frac<br />
{1}<br />
{\sigma \sqrt{2\pi}}<br />
e^{-(x^2/2\sigma^2 )}<br />
    is equal to

    \frac{3}{8}\pi ^{-1/2}\sigma ^{-5}.
    Can anybody help me here?
    Many thanks

    I can get the second derivative but im not sure how to square it or what to do with it once its squared.
    Last edited by mr fantastic; August 28th 2009 at 03:08 AM. Reason: Merged posts
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  2. #2
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    just do successive integrations by parts (2).
    Last edited by mr fantastic; August 28th 2009 at 03:10 AM.
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  3. #3
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    Quote Originally Posted by Moo View Post
    and are you sure it's not

    f(x)=\frac<br />
{1}<br />
{\sigma \sqrt{2\pi}}<br />
e^{-(x^2/2\sigma^{\color{red}2} )}<br />

    ?

    as defunkt mentioned, it should be bounded. if so, just do successive integrations by parts (2).
    #
    Could you help me with what I am supposed to Integrate by parts?
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  4. #4
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    Hi. You can integrate \int x^n e^{-ax^2}dx=\int x^{n-1}(x e^{-ax^2}dx) right? Just let u=x^{n-1} and dv=xe^{-ax^2} and then keep re-integrating by parts until you reduce the power on x to one.

    Above, you have f(x)=ae^{-bx^2} right? Surely you can take the second derivative to obtain f''=2abe^{-bx^2}(2bx^2-1). Now, that's no problem to square it and then expand out all the terms. You'll end up with terms like c_1 e^{-c_2 x^2}, c_3 x^2 e^{-c_2 x^2} and c_4 x^4 e^{-c_2 x^2} which you can then integrate by parts like above and solve.
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  5. #5
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    Thanks for the help but the integration by parts is gettin out of control for me. Thanks anyway.
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  6. #6
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    Quote Originally Posted by markrvr View Post
    Thanks for the help but the integration by parts is gettin out of control for me. Thanks anyway.
    Moo said two and she's smarter than me so maybe there's a better way to do this with just two integrations by parts.
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  7. #7
    Moo
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    And Moo forgot the product rule of differentiation

    But basically for each term of the integral, you need at most 2 integrations by parts. And since there are 3 of them, that's surely not a very nice thing to do.

    I'm thinking on a probabilistic way of solving it, but I can't seem to find it.
    How did you come up with this integral ?
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  8. #8
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    Quote Originally Posted by markrvr View Post
    Show that

    \int_{- \infty}^{+ \infty} (f''(x))^2 \, dx

    where

     <br />
f(x)=\frac<br />
{1}<br />
{\sigma \sqrt{2\pi}}<br />
e^{-(x^2/2\sigma^2 )}<br />
    is equal to

    \frac{3}{8}\pi ^{-1/2}\sigma ^{-5}.
    Can anybody help me here?
    Many thanks
    Obviously
     f'=-\frac{x}{\sigma^2} f \text{, } f''=\frac{x^2-\sigma^2}{\sigma^4} f

    I=\int_{-\infty}^{+\infty} [f''(x)]^2 \, dx=\int_{-\infty}^{+\infty} \frac{(x^2-\sigma^2)^2}{\sigma^8} f^2 \ dx=\frac{1}{\pi \sigma^{10}} \int_0^{+\infty} (x^4-2 \sigma^2 x^2 +\sigma^4) \ e^{-\frac{x^2}{\sigma^2}} \ dx

     \text{Let } u=\frac{x^2}{\sigma^2} \text{, then } I= \frac{1}{2 \pi \sigma^5} \int_0^{+\infty} (u^{\frac{3}{2}}-2u^{\frac{1}{2}} +u^{-\frac{1}{2}}) \ e^{-u} \ du = \frac{1}{2 \pi \sigma^5} \left[\Gamma(\tfrac{5}{2})-2\Gamma(\tfrac{3}{2})+\Gamma(\tfrac{1}{2})\right]=\frac{3}{8}\pi^{-1/2}\sigma ^{-5}
    Last edited by mr fantastic; September 18th 2009 at 07:34 AM. Reason: Restored reply
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  9. #9
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    Quote Originally Posted by luobo View Post
    Obviously
     f'=-\frac{x}{\sigma^2} f \text{, } f''=\frac{x^2-\sigma^2}{\sigma^4} f

    I=\int_{-\infty}^{+\infty} [f''(x)]^2 \, dx=\int_{-\infty}^{+\infty} \frac{(x^2-\sigma^2)^2}{\sigma^8} f^2 \ dx=\frac{1}{\pi \sigma^{10}} \int_0^{+\infty} (x^4-2 \sigma^2 x^2 +\sigma^4) \ e^{-\frac{x^2}{\sigma^2}} \ dx

     \text{Let } u=\frac{x^2}{\sigma^2} \text{, then } I= \frac{1}{2 \pi \sigma^5} \int_0^{+\infty} (u^{\frac{3}{2}}-2u^{\frac{1}{2}} +u^{-\frac{1}{2}}) \ e^{-u} \ du = \frac{1}{2 \pi \sigma^5} \left[\Gamma(\tfrac{5}{2})-2\Gamma(\tfrac{3}{2})+\Gamma(\tfrac{1}{2})\right]=\frac{3}{8}\pi^{-1/2}\sigma ^{-5}
    Well thanks for that! Could you just tell me why the bound of the integral changes from -infinity to zero on the second line and on the third line how you go from u to gamma and what happens to the e^-u, also could you tell me how you get from the part with the gammas in to the answer. Thanks very much.
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  10. #10
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    Quote Originally Posted by markrvr View Post
    Well thanks for that! Could you just tell me why the bound of the integral changes from -infinity to zero on the second line and on the third line how you go from u to gamma and what happens to the e^-u, also could you tell me how you get from the part with the gammas in to the answer. Thanks very much.
    The integrand is an even function (so just integrate from zero to infinity and multiply by two), and the Gamma function is defined as:

    \Gamma(x)=\int_0^{\infty} t^{x-1} e^{-t}dt so therefore the integral:

    \int_0^{\infty} u^{3/2}e^{-u}du=\int_0^{\infty} t^{5/2-1}e^{-t}dt=\Gamma(5/2)

    Also, check out the formula for \Gamma(n+1/2) to get the last part.

    . . . thanks Luobo.
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  11. #11
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    Hi,
    How did you get from the part with sigma to the 10 to the part with sigma to the 5? In other words how did tou get from x's and sigmas to the u's?
    And at the risk of sounding cheeky could you do this for

    \int_{- \infty}^{+ \infty} (f^{(4)}(x))^2 \, dx

    Thanks Very Much
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  12. #12
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    Quote Originally Posted by luobo View Post
    Obviously
     f'=-\frac{x}{\sigma^2} f \text{,   } f''=\frac{x^2-\sigma^2}{\sigma^4} f
    No, that is not "obvious" nor even correct. The derivative of \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}} is -\frac{x}{\sigma^3\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}} and the second derivative is -\frac{1}{\sigma^3\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}+ \frac{x}{2\sigma^5\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}.

    I=\int_{-\infty}^{+\infty} [f''(x)]^2 \, dx=\int_{-\infty}^{+\infty} \frac{(x^2-\sigma^2)^2}{\sigma^8} f^2 \ dx=\frac{1}{\pi \sigma^{10}} \int_0^{+\infty} (x^4-2 \sigma^2 x^2 +\sigma^4) \ e^{-\frac{x^2}{\sigma^2}} \ dx

     \text{Let } u=\frac{x^2}{\sigma^2} \text{, then } I= \frac{1}{2 \pi \sigma^5} \int_0^{+\infty} (u^{\frac{3}{2}}-2u^{\frac{1}{2}} +u^{-\frac{1}{2}}) \ e^{-u} \ du = \frac{1}{2 \pi \sigma^5} \left[\Gamma(\tfrac{5}{2})-2\Gamma(\tfrac{3}{2})+\Gamma(\tfrac{1}{2})\right]=\frac{3}{8}\pi^{-1/2}\sigma ^{-5}
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  13. #13
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    Quote Originally Posted by HallsofIvy View Post
    No, that is not "obvious" nor even correct. The derivative of \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}} is -\frac{x}{\sigma^3\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}} and the second derivative is -\frac{1}{\sigma^3\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}+ \frac{x}{2\sigma^5\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}.
    Is the rest of it correct though?
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  14. #14
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    Mathematica returns that reported by Luobo, that is:

    f^{''}=\frac{x^2-\sigma^2}{\sqrt{2\pi} \sigma^5} e^{-\frac{x^2}{2\sigma^2}}
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  15. #15
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    Originally Posted by luobo
    Obviously
    Quote Originally Posted by HallsofIvy View Post
    No, that is not "obvious" nor even correct. The derivative of \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}} is -\frac{x}{\sigma^3\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}} and the second derivative is -\frac{1}{\sigma^3\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}+ \frac{x}{2\sigma^5\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}.
    Your result of the second-order derivative is incorrect.

    Here is my approach, in more detail. Start from
    <br />
f'=-\frac{x}{\sigma^2} f \text{ (Eqn. 1)}<br />
    Use (Eqn. 1) recursively, you can find  f'', f''' \text{ and so on}

    For example,
    <br />
f''=-\frac{x}{\sigma^2} f'-\frac{1}{\sigma^2} f=\frac{x^2-\sigma^2}{\sigma^4} f<br />
    <br />
f'''=\frac{x^2-\sigma^2}{\sigma^4} f' + \frac{2x}{\sigma^4} f = \frac{3x\sigma^2-x^3}{\sigma^6}f<br />
    Last edited by mr fantastic; September 18th 2009 at 07:36 AM. Reason: Restored original post
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