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Consider the following function \begin{align} g(x)=e^{-x/4}\int_0^{x/4}e^y\vartheta(2\textstyle\sqrt{y/\pi})dy, \end{align} where $\vartheta$ is the theta function, given by $$ \vartheta(x)=\sum_{n\in\mathbb{Z}}e^{-\pi(nx)^2} $$ which satisfies $\vartheta(1/x)=x\vartheta(x)$. In this answer, it was said that, for small $x$, we have $g(x)=\sqrt\pi D^+(\sqrt{x}/2)+O(xe^{-\pi^2/x})$, where $D^+$ is the Dawson function, given by \begin{align} D^+(z)=e^{-z^2}\int_0^z e^{t^2}dt. \end{align} It was then said that a less accurate estimate to $g$ can be given by $g(x)=\sqrt{\pi x}/2+O(x^{3/2})$.

I am struggling to understand both of these approximations. Any ideas?

sam wolfe
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1 Answers1

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If I'm not mistaken, I think we have $$g(x)=\sqrt\pi D^+(\sqrt{x}/2)+O\bigg(\color{red}{x^{3/2}} e^{-\pi^2/x}\bigg)$$

Setting $t=\sqrt y$, we get

$$\begin{align}g(x)&=e^{-x/4}\int_0^{x/4}e^y\vartheta(2\textstyle\sqrt{y/\pi})dy \\\\&=e^{-x/4}\int_0^{\sqrt x/2}e^{t^2}\vartheta(2\textstyle\sqrt{t^2/\pi})2tdt \\\\&=e^{-x/4}\int_0^{\sqrt x/2}\sqrt{\pi}e^{t^2}\vartheta(2t/\sqrt{\pi})2t/\sqrt{\pi}dt \\\\&=\sqrt{\pi}e^{-x/4}\int_0^{\sqrt x/2}e^{t^2}\vartheta\bigg(\frac{\sqrt{\pi}}{2t}\bigg)dt \\\\&=\sqrt{\pi}e^{-x/4}\int_0^{\sqrt x/2}e^{t^2}\sum_{n\in\mathbb{Z}}e^{-\frac{n^2\pi^2}{4t^2}}dt \\\\&=\sqrt{\pi}e^{-x/4}\int_0^{\sqrt x/2}e^{t^2}\bigg(1+2\sum_{n\ge 1}e^{-\frac{n^2\pi^2}{4t^2}}\bigg)dt \\\\&=\sqrt\pi D^+(\sqrt{x}/2)+2\sqrt{\pi}e^{-x/4}\underbrace{\int_0^{\sqrt x/2}\sum_{n\ge 1}e^{t^2-\frac{n^2\pi^2}{4t^2}}dt}_{G(x)}\end{align}$$

Here, we have $$G(0)=0\qquad\text{and}\qquad G'(x)=\sum_{n\ge 1}\frac{e^{x/4-n^2\pi^2/x}}{4\sqrt x}$$ Since $(\text{erf}(x))'=\frac{2}{\sqrt{\pi}}e^{-x^2}$, we have $$\bigg(\text{erf}\bigg(\frac{n\pi}{\sqrt x}\pm\frac{i\sqrt x}{2}\bigg)\bigg)'=e^{x/4-n^2\pi^2/x}(-1)^{n+1}\bigg(\frac{n\sqrt{\pi}}{x^{3/2}}\mp\frac{i}{2\sqrt{\pi x}}\bigg)$$

Using this, we obtain $$G(x)=\sum_{n\ge 1}\frac{(-1)^ni\sqrt{\pi}}{4}\bigg(\text{erf}\bigg(\frac{n\pi}{\sqrt{x}}-\frac{ i\sqrt x}{2}\bigg) - \text{erf}\bigg(\frac{n\pi}{\sqrt x}+\frac{i\sqrt x}{2}\bigg)\bigg) $$

Using $$\text{erf}(X)=1-\frac{e^{-X^2}}{\sqrt{\pi}X}+O\bigg(\frac{e^{-X^2}}{X^3}\bigg)$$ we have, for small $x$,

$$\begin{align}G(x)&= \sum_{n\ge 1}\frac{(-1)^ni\sqrt{\pi}}{4} \bigg(1-\frac{e^{-(\frac{n\pi }{\sqrt x}-\frac{i\sqrt x}{2})^2}}{\sqrt{\pi}(\frac{n\pi }{\sqrt x}-\frac{i\sqrt x}{2})} -\bigg(1-\frac{e^{-(\frac{n\pi }{\sqrt x}+\frac{i\sqrt x}{2})^2}}{\sqrt{\pi}(\frac{n\pi }{\sqrt x}+\frac{i\sqrt x}{2})} \bigg)\bigg) \\\\&=\sum_{n\ge 1} \frac{(-1)^ni\sqrt{\pi}}{4} \bigg(-\frac{e^{-\frac{n^2\pi^2}{x}+n\pi i+\frac{x}{4}}}{\sqrt{\pi}(\frac{n\pi }{\sqrt x}-\frac{i\sqrt x}{2})}+\frac{e^{-\frac{n^2\pi^2}{x}-n\pi i+\frac{x}{4}}}{\sqrt{\pi}(\frac{n\pi }{\sqrt x}+\frac{i\sqrt x}{2})} \bigg) \\\\&=\sum_{n\ge 1} \frac{(-1)^ni\sqrt{\pi}}{4} \bigg(-\frac{(-1)^ne^{-\frac{n^2\pi^2}{x}+\frac{x}{4}}}{\sqrt{\pi}(\frac{n\pi }{\sqrt x}-\frac{i\sqrt x}{2})}+\frac{(-1)^ne^{-\frac{n^2\pi^2}{x}+\frac{x}{4}}}{\sqrt{\pi}(\frac{n\pi }{\sqrt x}+\frac{i\sqrt x}{2})} \bigg) \\\\&=\sum_{n\ge 1} \frac{i\sqrt{\pi}}{4} \bigg(-\frac{e^{x/4-n^2\pi^2/x}}{\sqrt{\pi}(\frac{n\pi }{\sqrt x}-\frac{i\sqrt x}{2})}+\frac{e^{x/4-n^2\pi^2/x}}{\sqrt{\pi}(\frac{n\pi }{\sqrt x}+\frac{i\sqrt x}{2})} \bigg) \\\\&=\sum_{n\ge 1}\frac{x^{3/2}}{x^2+4n^2\pi^2}e^{x/4-n^2\pi^2/x} \\\\&=\sum_{n\ge 1}\bigg(\frac{x^{3/2}}{4n^2\pi^2}+O\bigg(x^{7/2}\bigg)\bigg)e^{x/4-n^2\pi^2/x}\end{align}$$ from which we have $$g(x)=\sqrt\pi D^+(\sqrt{x}/2)+O\bigg(x^{3/2} e^{-\frac{\pi^2}{x}}\bigg)\tag1$$

As commented by Mariusz Iwaniuk, using $$D_+(z)=\frac12\sum_{n=0}^\infty\frac{(-1)^n n!}{(2n+1)!}(2z)^{2n+1}$$ we have $$\sqrt\pi D^+(\sqrt{x}/2)=\frac{\sqrt{\pi}}2\sum_{n=0}^\infty\frac{(-1)^n n!}{(2n+1)!}(\sqrt{x})^{2n+1}=\sqrt{\pi x}/2+O(x^{3/2})\tag2$$

It follows from $(1)(2)$ that $$g(x)=\sqrt{\pi x}/2+O(x^{3/2})$$

mathlove
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  • Thanks! I was just wondering, is there really the need to call the derivative of the erf function and complex numbers to show the first estimate? Is there a way to shortcut those steps? – sam wolfe Sep 07 '23 at 09:17