Integrate second derivative squared.

**markrvr** · August 25th 2009, 02:02 AM

Show that

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

where

$ f(x)=\frac {1} {\sigma \sqrt{2\pi}} e^{-(x^2/2\sigma^2 )} $
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.

**Moo** · August 25th 2009, 02:28 AM

just do successive integrations by parts (2).

**markrvr** · August 28th 2009, 02:07 AM

Originally Posted by Moo

and are you sure it's not

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

?

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?

**shawsend** · August 28th 2009, 08:23 AM

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.

**markrvr** · August 29th 2009, 07:29 AM

Thanks for the help but the integration by parts is gettin out of control for me. Thanks anyway.

**shawsend** · August 29th 2009, 08:51 AM

Originally Posted by markrvr

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.

**Moo** · August 30th 2009, 04:58 AM

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 ?

**luobo** · August 30th 2009, 06:54 AM

Originally Posted by markrvr

Show that

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

where

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

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

Obviously
$f'=-\frac{x}{\sigma^2} f \text{, } f''=\frac{x^2-\sigma^2}{\sigma^4} f$

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

$\text{Let } u=\frac{x^2}{\sigma^2} \text{, then } I= \frac{1}{2 \pi \sigma^5} \int_0^{+\infty} (u^{\frac{3}{2}}-2u^{\frac{1}{2}} +u^{-\frac{1}{2}}) \ e^{-u} \ du$ $= \frac{1}{2 \pi \sigma^5} \left[\Gamma(\tfrac{5}{2})-2\Gamma(\tfrac{3}{2})+\Gamma(\tfrac{1}{2})\right]=\frac{3}{8}\pi^{-1/2}\sigma ^{-5}$

**markrvr** · August 31st 2009, 12:19 PM

Originally Posted by luobo

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.

**shawsend** · August 31st 2009, 01:06 PM

Originally Posted by markrvr

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.

**markrvr** · September 1st 2009, 01:14 AM

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

**HallsofIvy** · September 1st 2009, 06:25 AM

Originally Posted by luobo

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 $-\frac{1}{\sigma^3\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}+ \frac{x}{2\sigma^5\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}$ .

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

$\text{Let } u=\frac{x^2}{\sigma^2} \text{, then } I= \frac{1}{2 \pi \sigma^5} \int_0^{+\infty} (u^{\frac{3}{2}}-2u^{\frac{1}{2}} +u^{-\frac{1}{2}}) \ e^{-u} \ du$ $= \frac{1}{2 \pi \sigma^5} \left[\Gamma(\tfrac{5}{2})-2\Gamma(\tfrac{3}{2})+\Gamma(\tfrac{1}{2})\right]=\frac{3}{8}\pi^{-1/2}\sigma ^{-5}$

**markrvr** · September 1st 2009, 08:02 AM

Originally Posted by HallsofIvy

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?

**shawsend** · September 1st 2009, 08:11 AM

Mathematica returns that reported by Luobo, that is:

$f^{''}=\frac{x^2-\sigma^2}{\sqrt{2\pi} \sigma^5} e^{-\frac{x^2}{2\sigma^2}}$

**luobo** · September 1st 2009, 03:43 PM

Originally Posted by luobo

Obviously

Originally Posted by HallsofIvy

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
$ f'=-\frac{x}{\sigma^2} f \text{ (Eqn. 1)} $
Use (Eqn. 1) recursively, you can find $f'', f''' \text{ and so on}$

For example,
$ f''=-\frac{x}{\sigma^2} f'-\frac{1}{\sigma^2} f=\frac{x^2-\sigma^2}{\sigma^4} f $
$ f'''=\frac{x^2-\sigma^2}{\sigma^4} f' + \frac{2x}{\sigma^4} f = \frac{3x\sigma^2-x^3}{\sigma^6}f $

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