Can one show this is equivalent to erf(x) by elementary means?

Question

According to the PDF at this link (eqn 12), the following is true: $$\text{erf}(x)=\frac{1}{\pi}\int_0^\infty e^{-t} \sin(2x\sqrt{t})\frac{dt}{t} $$

I would like to use this identity as part of a presentation to my students (the integral can be derived in an intuitive way for a heat diffusion problem). If need be I can simply assert "and this is known to be equivalent to the error function" but I'd rather prove it for them. However since the group are not mathematicians I would need to be able to show it using elementary methods; for example integration by parts is fine, but contour integration is not.

I have tried and failed to show the equivalence. Any suggestions welcome.

Answer 1

By the substitution $\sqrt{t}\rightarrow \pi u$, we get \begin{align} 2\frac{1}{\pi}\int^\infty_0 e^{-\pi^2u^2}\frac{\sin(2\pi x u)}{u}\ du = \int^\infty_{-\infty} e^{-\pi^2u^2}\frac{\sin(2\pi x u)}{\pi u}\ du. \end{align}

Apply Plancherel's theorem, which states \begin{align} \int^\infty_{-\infty} f(x)\overline{g(x)}\ dx = \int^\infty_{-\infty} \hat f(\xi)\overline{\hat g(\xi)}\ d\xi, \end{align} then it follows \begin{align} \int^\infty_{-\infty} e^{-\pi^2u^2}\frac{\sin(2\pi x u)}{\pi u}\ du=&\ \frac{1}{\sqrt{\pi}}\int^\infty_{-\infty} e^{-t^2}\chi_{[-x, x]}(t)\ dt = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2}\ dt =\operatorname{erf}(x). \end{align}

I will leave it as an exercise for the reader to check

\begin{align} \int^\infty_{-\infty} e^{-\pi^2 x^2}e^{-2\pi i x\xi}\ dx = \frac{1}{\sqrt{\pi}}e^{-\xi^2}. \end{align}

Hint: Complete the square. Formally ignored the complex number in the square and perform a $u$-substitution.

Answer 2

\begin{align} \tag{1a} \int\limits_{0}^{\infty} \mathrm{e}^{-t} \frac{\sin(2x\sqrt{t})}{t} dt &= x \int\limits_{-1}^{1} \int\limits_{0}^{\infty} \mathrm{e}^{-t} \mathrm{e}^{i2x\theta \sqrt{t}} \frac{1}{\sqrt{t}} dt d\theta \\ \tag{1b} &= 2x \int\limits_{-1}^{1} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}+i2x\theta y} dy d\theta \\ \tag{1c} &= 2x \int\limits_{-1}^{1} \mathrm{e}^{-x^{2} \theta ^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-(y-ix\theta)^{2}} dy d\theta \\ \end{align}

Notes:

a. Use $$\frac{\sin(\phi)}{\phi} = \frac{1}{2}\int\limits_{-1}^{1} \mathrm{e}^{i\phi w} dw$$

b. $\sqrt{t} = y$

c. Complete the square.

Evaluate \begin{align} \int \mathrm{e}^{-(y-ix\theta)^{2}} dy &= \int \mathrm{e}^{-z^{2}} dz \\ &= \frac{\sqrt{\pi}}{2} \mathrm{erf}(z) = \frac{\sqrt{\pi}}{2} \mathrm{erf}(y-ix\theta) \end{align} here we let $z = y-ix\theta $. Applying limits to the integral yields \begin{align} \int\limits_{0}^{\infty} \mathrm{e}^{-(y-ix\theta)^{2}} dy &= \frac{\sqrt{\pi}}{2} \mathrm{erf}(y-ix\theta) \Big|_{0}^{\infty} \\ &= \frac{\sqrt{\pi}}{2} [1 + i \mathrm{erfi}(x\theta)] \tag{2} \end{align}

Substituting equation 2 into equation 1c yields two integrals. Dropping constants we have first \begin{align} \int\limits_{-1}^{1} \mathrm{e}^{-x^{2} \theta ^{2}} d\theta &= \frac{1}{x} \int \mathrm{e}^{-w^{2}} dw \\ &= \frac{\sqrt{\pi}}{2x} \mathrm{erf}(w) \\ &= \frac{\sqrt{\pi}}{2x} \mathrm{erf}(x\theta) \Big|_{-1}^{1} \\ &= \frac{\sqrt{\pi}}{x} \mathrm{erf}(x) \tag{3} \end{align} secondly, we have \begin{equation} \int\limits_{-1}^{1} \mathrm{e}^{-x^{2} \theta ^{2}} \mathrm{erfi}(x\theta) d\theta = 0 \end{equation} noting that the imaginary error function is an odd function.

Putting the pieces together yields \begin{align} \int\limits_{0}^{\infty} \mathrm{e}^{-t} \frac{\sin(2x\sqrt{t})}{t} dt &= 2x \int\limits_{-1}^{1} \mathrm{e}^{-x^{2} \theta ^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-(y-ix\theta)^{2}} dy d\theta \\ &= 2x \frac{\sqrt{\pi}}{2} \frac{\sqrt{\pi}}{x} \mathrm{erf}(x) \\ &= \pi \mathrm{erf}(x) \end{align}