Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello,

Here is an interesting problem. It looks elementary, but it has taken me some efforts without solving it. Let

$$ h(x) = e^{x^2/2} \Phi(x),\qquad \text{with}\quad \Phi(x):=\int_{-\infty}^x \frac{e^{-y^2/2}}{\sqrt{2\pi}} dy. $$

The question is whether the function $h(x)$ is monotone increasing over $R$? Are there some work dealing with such function?

It seems a quite easy problem. By taking the first derivative, we need to prove that

$$ h(x)' = h(x) x + \frac{1}{\sqrt{2\pi}} \ge 0. $$ which again, not obvious (for $x<0$). Some facts, that might be useful, are:

$$ \lim_{x\rightarrow -\infty} h(x) =0, \quad \lim_{x\rightarrow -\infty} h(x)' =0. $$

Thank you very much for any hints!

Anand

share|improve this question
1  
Dawson's integral seems closely related mathworld.wolfram.com/DawsonsIntegral.html –  Aaron Hoffman Oct 3 '11 at 12:58
    
@Aaron Hoffman, thank you very much. I also noticed latter that it is closely related to this special function. Actually, in Chapter 41 of the book: books.google.com/… –  Anand Oct 4 '11 at 8:45
add comment

3 Answers

up vote 22 down vote accepted

We can write $h(x)=\frac 1{\sqrt{2\pi}}\int_{-\infty}^x \exp\left(\frac{x^2-y^2}2\right)dy$. Now put $t=x-y$. We get \begin{align} h(x)&=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}\exp\left(\frac{x^2-(x-t)^2}2\right)dt\\\ &=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}\exp\left(xt-\frac{t^2}2\right)dt. \end{align} We can differentiate under the integral thanks to the dominated convergence theorem. We get $$h'(x)=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}t\exp\left(xt-\frac{t^2}2\right)dt\geq 0.$$ Added later: we don't need to diffentiate. If $x_1\leq x_2$ then for $t\geq 0$ we have $e^{tx_1}\leq e^{tx_2}$ therefore $ h(x_1)\leq h(x_2) $.

share|improve this answer
4  
Nice argument... –  Igor Rivin Oct 3 '11 at 10:09
    
@Davide Giraudo, Thanks a lot! :-) –  Anand Oct 3 '11 at 10:11
add comment

This is just an alternative argument to Davide's nice one.

First, note that $h' = e^{x^2/2}(x \Phi + \Phi')$.

Since $\Phi'' = -x \Phi'$, monotonicity of the integral yields $$ x \Phi(x) \geq \int_{-\infty}^x u \Phi'(u) \mathrm{d}u = -\Phi'(x). $$ So, $h' \geq 0$, and we are done.

share|improve this answer
    
Thanks Cardinal. It is an equally nice solution. :-) –  Anand Oct 17 '11 at 20:29
add comment

I haven't actually done the computation, but it seems to me that integrating the $\Phi(x)$ term by parts ad nauseam, you get a nice power series for $h(x).$

EDIT @Davide's argument is obviously the complete answer to the question as asked, but just as a coda, the series for $h(x)$ is quite cute:

In the odd part, the coefficients of $x^{2k+1}$ is $1/p(k)$ where $p(k)$ is the product of the first $k$ odd integers, while in the even part, the coefficient of $x^{2k}$ is $\sqrt{\pi/2}/q(k),$ where $q(k)$ is the product of the first $k$ even integers.

share|improve this answer
    
Thanks Igor Rivin, I am afraid of series, especially if you multiply two series, one is of $e^{-x^2}$ and the other is of $\Phi(x)$. Since Davide has given a nice solution, I am very content now. Thanks again.:-) –  Anand Oct 3 '11 at 10:18
1  
I am glad you are content, but you might want to get over that series phobia (although the series is not useful for answering your question, it seems...) –  Igor Rivin Oct 3 '11 at 10:22
    
Thanks Igor Rivin. You are right. :-) –  Anand Oct 3 '11 at 10:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.