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I am studying Dawson's function : $\displaystyle F : x \mapsto e^{-x^2}\int_0^x e^{t^2} dt$.

I would like to prove that $F$ attains a maximum at a certain value $x_0 \in (0,1)$, and is increasing over $[0,x_0]$ and decreasing over $[x_0, +\infty)$.

The only thing I managed to prove about this function is that $\displaystyle\lim_{x \rightarrow +\infty} F(x)=0$, which gives the existence of a maximum over $[0,+\infty)$, but does not help to determine the variations of $F$.

I also know that $F$ is solution of the differential equation $y'+2xy=1$, but I cannot see how it could help.

So the question is : how to determine the variations of $F$ ? (and bonus : how to prove that $x_0 <1$ ?)


EDIT : Thanks to @NinadMunshi's answer, I understand now why there exists $x_0 \in (0,1)$ such that $F'(x_0)=0$. But I still don't understand how to prove that $x_0$ is the only root of $F'$ over $[0,+\infty)$ (which would solve my initial question) Does anybody know how to tackle this question ?

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3 Answers 3

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Let $x \geq 0$. Then

\begin{align*} 2x F(x) &= 2x e^{-x^2} \int_{0}^{x} e^{t^2} \, \mathrm{d}t \\ &= e^{-x^2} \int_{0}^{1} 2x^2 e^{x^2 s^2} \, \mathrm{d}s \tag{$t = xs$} \\ &= e^{-x^2} \biggl( \biggl[ \frac{e^{x^2 s^2} - 1}{s} \biggr]_{s=0}^{s=1} + \int_{0}^{1} \frac{e^{x^2s^2} - 1}{s^2} \, \mathrm{d}s \biggr) \\ &= 1 - e^{-x^2} + e^{-x^2} G(x), \end{align*}

where $G(x)$ is defined by

$$ G(x) = \int_{0}^{1} \frac{e^{x^2s^2} - 1}{s^2} \, \mathrm{d}s. $$

So it follows that

\begin{align*} F'(x) = 1 - 2x F(x) = e^{-x^2}( 1 - G(x) ). \end{align*}

However, it is clear that $G(0) = 0$ and $G(x)$ is strictly increasing on $x \geq 0$ with $G(x) \to \infty$ as $x \to \infty$. So, there exists a unique positive solution $x = x_0$ of the equation

$$ G(x) = 1, $$

and we have

$$ \begin{cases} F'(x) > 0 \iff x < x_0, \\ F'(x) = 0 \iff x = x_0, \\ F'(x) < 0 \iff x > x_0. \end{cases} $$

This proves that $F(x)$ attains a unique maximum on $[0, \infty)$.


Addendum. Using the inequality $e^x > 1 + x$ for $x > 0$, we know that

$$ G(1) = \int_{0}^{1} \frac{e^{s^2} - 1}{s^2} \, \mathrm{d}s > \int_{0}^{1} \, \mathrm{d}s = 1. $$

This implies $x_0 < 1$ as required.

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  • $\begingroup$ Thank you very much, that is perfect ! $\endgroup$
    – Henry
    Commented May 31 at 7:16
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Hint (too long for a comment)

There are two points to recall $$F(x)=\frac{\sqrt{\pi }}{2}\sum_{n=0}^\infty (-1)^n\, \frac{x^{2 n+1}}{\Gamma \left(n+\frac{3}{2}\right)}$$ and $$F'(x)=1-2 x F(x)=\sqrt{\pi }\sum_{n=0}^\infty (-1)^n\,\frac{x^{2 n}}{\Gamma \left(n+\frac{1}{2}\right)}$$ Build the $[2n,2n]$ Padé approximant $P_n$ of $F'(x)$ such as $$P_2=\frac {1-\frac{10 }{7}x^2+\frac{32 }{105}x^4 }{1+\frac{4 }{7}x^2+\frac{4}{35}x^4 }$$ To give an idea of the accuracy, compute numerically the infinite norm $$\Phi_2=\int_0^1 \Big(F'(x)-P_2 \Big)^2\,dx=4.22213\times 10^{-7}$$

If you are ready to work with cubic equations in $x^2$ $$\Phi_3=\int_0^1 \Big(F'(x)-P_3 \Big)^2\,dx=2.14162\times 10^{-11}$$ that is to say that you can have a good approximation of $x_*$ such that $F'(x_*)=0$.

Yous could do the same kind of work with $$F''(x)=2\left(2 x^2-1\right) F(x)-2 x$$ and find the inflection point.

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    $\begingroup$ Thanks for the answer ! However, I can't understand why $F'$ should have an unique root over $(0,+\infty)$. Am I missing something obvious ? $\endgroup$
    – Henry
    Commented May 30 at 9:44
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From your differential equation we know that the maximum occurs when $2xy=1$. On the interval $[0,1]$ we have that

$$0\cdot F(0) = 0$$

$$2\cdot F(1) = \frac{2}{e}\int_0^1 e^{t^2}dt > 1$$

So the solution where it equals $1$ must exist on the interval by intermediate value theorem.

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  • $\begingroup$ Thanks for the answer. However, if I undestand it correctly, it only proves that the derivative of $F$ vanishes at a certain point of the interval $(0,1)$. My question was rather about the global variations of $F$ over $(0,+\infty)$. Do you know how to prove that $F$ is increasing and then decreasing on this interval ? $\endgroup$
    – Henry
    Commented May 30 at 9:31
  • $\begingroup$ @Henry intermediate value theorem still applies. When you know the only root of $y'$ occurs at a single point, what can you say about the sign of $y'$ when $x>x_0$? $x<x_0$? $\endgroup$ Commented May 30 at 9:34
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    $\begingroup$ Precisely, I don't know why $y'$ has an only root over $(0,+\infty)$... Am I missing something obvious ? $\endgroup$
    – Henry
    Commented May 30 at 9:35

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