
#1
Feb1311, 03:52 PM

P: 590

Hi,
I have this derivation, and I am not sure how to derive it: [tex]\frac{d}{dp(x)}\int_{\infty}^{\infty}p(x)\,log(p(x))\,dx[/tex] I mean, what to do with the integral? Another thing, how to integrate: [tex]\int_{\infty}^{\infty}e^{x^2}\,dx[/tex] Thanks in advance 



#2
Feb1311, 04:40 PM

P: 751

For the first one, read here:
http://en.wikipedia.org/wiki/Differe..._integral_sign For the second one, square the thing and use Fubini's theorem. Then square root the answer at the end. 



#3
Feb1311, 07:43 PM

P: 590





#4
Feb1411, 05:51 PM

P: 751

Differentiation QuestionI^{2} = [tex](\int_{\infty}^{\infty}e^{x^2}\,dx)(\int_{\infty}^{\infty}e^{x^2}\,dx)[/tex] x is a dummy variable of integration. You can change it to "y" (or your favorite letter) [tex](\int_{\infty}^{\infty}e^{x^2}\,dx)(\int_{\infty}^{\infty}e^{y^2}\,dy)[/tex] This is similar to a double integral where it got split up because the first integrand was independent of the other variable. You can put them back together (you should know what Fubini's theorem says). Switch to polar coordinates and it becomes a simple integral over the entire rtheta plane. Square root the answer at the end (since you just calculated the square of the answer, remember you squared the integral) Eat cookie 



#5
Feb1411, 06:20 PM

Sci Advisor
HW Helper
P: 11,863





#6
Feb1511, 09:01 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Good point bigubau! The others were thinking of
[tex]\int_{\infty}^{\infty} e^{x^2}dx[/tex] S David, you seem to have difficulty distinguishing between a "definite" integral and "indefinite" integral. The first problem was also a "definite" integral and so, also as bitubau said, a constant. It's derivative is 0. The others were thinking of differentiating the "indefinite integral" [tex]\frac{d}{dp(x)}\int p(x)\,log(p(x))\,dx[/tex] 



#7
Feb1511, 09:57 AM

P: 590

[tex]\underset{p(x)}{\text{max}}\int_{\infty}^{\infty}p(x)\log_2[p(x)]\,dx[/tex] subject to: [tex]\int_{\infty}^{\infty}p(x)\,dx=1[/tex] 



#8
Feb1511, 02:55 PM

HW Helper
P: 1,391

That is, I think what you want is [tex]\frac{\delta}{\delta p(y)}\int_{\infty}^{\infty}p(x)\,log(p(x))\,dx[/tex] The usual rules of regular calculus usually apply, but note [tex]\frac{\delta f(x)}{\delta f(y)} = \delta(xy)[/tex] where [itex]\delta(xy)[/itex] is the dirac delta. In doing this you will not get zero as an answer. 


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