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You are given the result that $$\int^{+\infty}_{-\infty} e^{-x^2} dx = \sqrt{\pi}$$

a. Use this result to find $$\int^{+\infty}_{-\infty} e^{-ax^2} dx$$ b. Use the above results to find $$\int^{+\infty}_{-\infty} x^2e^{-ax^2} dx$$ [Hint: Consider $\frac{d}{d\alpha} \int^{+\infty}_{-\infty}e^{-ax^2}dx$]

c. Use the above results to to show that $$P(x,t)=\frac{1}{\sqrt{4 \pi D t}} \exp \left( -\frac{x^2}{4Dt}\right) $$ is a normalized distribution.

(Note: I'm still learning - this problem is somewhat advanced for my level, so if someone could write out an explicit and complete solution, that would be the most helpful answer for me.)

As of now, @matt has successfully helped me to understand the solutions for (a) and (b) in his response below; but I still need to have the solution for (c) explained to me. I do not understand what needs to be done. for (c).

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I believe first integral should be: $\int^{+\infty}_{-\infty} e^{-x^2} dx = \mathbf{\sqrt{\pi}}$ not $\sqrt{x}$. – matt Nov 16 '11 at 10:51
@Davide Giraudo Yes, you and matt are both correct - the typo has been corrected. (And yes, this is a homework problem.) – ptrcao Nov 16 '11 at 10:57
@Davide Giraudo What substitution am I supposed to be using? – ptrcao Nov 16 '11 at 10:59
To add to Davide Giraudo's excellent suggestion about integrating by parts for the second question, you should first write down the derivative of $e^{-ax^2}$ which will help you in writing the integrand in the form $u\,dv$ that is needed for integration by parts: you want $dv$ to be something whose antiderivative $v$ is known to you. – Dilip Sarwate Nov 16 '11 at 11:42
I hope you have looked at wikipedia – picakhu Dec 2 '11 at 21:36
up vote 10 down vote accepted

a. We make the substitution $t=\sqrt{a}x$, $a>0$, thus: $$ \int_{-\infty}^{+\infty}e^{-ax^2}dx =\lim_{\substack{n\rightarrow\infty\\m\rightarrow\infty}}\int_{-n}^{+m}e^{-ax^2}dx =\lim_{\substack{n\rightarrow\infty\\m\rightarrow\infty}}\int_{-n\sqrt{a}}^{+m\sqrt{a}}e^{-t^2}\frac{dt}{\sqrt{a}} =\frac1{\sqrt{a}}\int^{+\infty}_{-\infty} e^{-x^2} dx = \sqrt{\frac{\pi}{a}} $$

b. We differentiating our result from part (a) with respect to $a$. $$ \frac{d}{da} \int^{+\infty}_{-\infty}e^{-ax^2}dx = \frac{d}{da}\left(\sqrt{\frac{\pi}{a}}\right) \quad\Rightarrow\quad \int^{+\infty}_{-\infty}\frac{d}{da}e^{-ax^2}dx = \frac{d}{da}\left(\sqrt{\frac{\pi}{a}}\right) $$

Note that moving $\frac{d}{da}$ inside the integral is justified since both $e^{-ax^2}$ and $\frac{d}{da}(e^{-ax^2})$ are continuous.

Simplifying gives: $$ -\int_{-\infty}^{+\infty}x^2e^{-ax^2}dx = -\frac{\sqrt{\pi}}{2a^{3/2}} \quad\Rightarrow\quad \int_{-\infty}^{+\infty}x^2e^{-ax^2}dx = \frac{\sqrt{\pi}}{2a^{3/2}} $$

c. We must show: (1) $P(x,t)$ is non-negative; and (2) $\int P(x,t)=1$. We establish (1) by noting that $\exp(-x^2/4DT)\geq 0$ for all $x\in\mathbb{R}$. To establish (2), observe the following result which follows from part (a) (we set $a=1/(4Dt)$): $$ \int_{-\infty}^{+\infty}P(x,t)\;dx =\frac{1}{\sqrt{4 \pi D t}} \int_{-\infty}^{+\infty} \exp \left( -\frac{x^2}{4Dt}\right)\;dx =\frac{1}{\sqrt{4 \pi D t}} \sqrt{\frac{\pi}{1/(4Dt)}} = 1 $$

EDIT: Added some clarification for part (c).

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Thank you, I have finally caught up with your solutions for (a.) and (b.) but I'm still trying to understand what is happening in your solution for (c.) I don't understand what we need to do to show that the function P(x,t) is a normalised distribution. If you or anyone can offer me some walkthrough commentary, that would be most helpful... – ptrcao Nov 26 '11 at 13:40
I would say that $\frac{1}{\sqrt{4 \pi D t}} \exp \left( -\frac{x^2}{4Dt}\right)$ is a probability distribution in that it is non-negative and its integral is $1$. matt has shown the second point, and the first is obvious. normalized is a slightly confusing word here, with more than one obvious meaning. – Henry Nov 26 '11 at 15:48
@ptrcao I unintentionally left out the dx. Thank you for letting me know I have edited appropriately. I agree with Henry on his definition. Wikipedia has the following definition for normalising constant: "A normalizing constant is a constant by which an everywhere non-negative function must be multiplied so the area under its graph is 1" ( In the context of (c) the normalizing constant is 1/(4Dt). – matt Nov 26 '11 at 21:58
@ptcao: Planet Math has the formal measure-theoretic definition. – Henry Nov 26 '11 at 22:14
@ptrcao: it is the measure for the space: on the real line the measure of the interval $[a,b]$ is usually the length, i.e. $b-a$. – Henry Nov 27 '11 at 0:04

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