# Variations of Dawson's function

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 ?

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.

• Thank you very much, that is perfect ! Commented May 31 at 7:16

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.

• Thanks for the answer ! However, I can't understand why $F'$ should have an unique root over $(0,+\infty)$. Am I missing something obvious ? Commented May 30 at 9:44

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.

• 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 ? Commented May 30 at 9:31
• @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 $x>x_0$? $x<x_0$? Commented May 30 at 9:34
• Precisely, I don't know why $y'$ has an only root over $(0,+\infty)$... Am I missing something obvious ? Commented May 30 at 9:35