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I would like ask how to compute the Fourier transform $F(k)$ of the following function:

$\exp\left[-\frac{\eta^2}{8\sigma (0,t)}\right]\text{erf}\left(\frac{\eta}{\sqrt{8\sigma(0,t)}}\right)$

It can be seen that differentiating $\text{erf}\left(\frac{\eta}{\sqrt{8\sigma(0,t)}}\right)$ is $\exp\left[-\frac{\eta^2}{8\sigma (0,t)}\right]$ times a constant. It may reduce to compute Fourier transform of $\frac{d F(\eta)^2}{d\eta}$, where $F(\eta)=\text{erf}\left(\frac{\eta}{\sqrt{8\sigma(0,t)}}\right)$. Could someone provide solution to the calculation? In case if it cannot be found explicitly, I would also like to know how to express its Fourier transform $F(k)$ in series expansion.

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  • $\begingroup$ What is $\sigma(0,t)$ supposed to mean? is $\eta$ the variable you're supposed to be taking the FT with respect to? $\endgroup$ – Batman Sep 9 '17 at 5:37
  • $\begingroup$ We do Fourier transformation on $\eta$ only. You can treat any function in terms of $t$ to be constant. It maybe not easy to get explicitfunction, getting expansion in terms of Fourier variable up to order 3 is fine. $\endgroup$ – will_cheuk Sep 9 '17 at 13:15
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Let be $$ f(\eta)=\exp\left[-\frac{\eta^2}{8\sigma (0,t)}\right]\text{erf}\left(\frac{\eta}{\sqrt{8\sigma(0,t)}}\right)=\mathrm e^{-a^2\eta^2}\text{erf}\left(a\eta\right) $$ where $a=\frac{1}{\sqrt{8\sigma(0,t)}}$ and $\mathrm{erf}(x)=\frac{2}{\sqrt\pi}\int_0^x\mathrm e^{-u^2}\mathrm d u$.

We have $$ \mathcal F\{f\}(\omega)=F(\omega)=\int_{-\infty}^\infty f(\eta)\,\mathrm e^{-i\omega\eta}\mathrm d\eta=\int_{-\infty}^\infty \mathrm e^{-a^2\eta^2}\text{erf}\left(a\eta\right)\,\mathrm e^{-i\omega\eta}\mathrm d\eta $$ Putting $\xi=a\eta$ we have $$ \begin{align} F(\omega)&=\frac{1}{a}\int_{-\infty}^\infty \text{erf}\left(\xi\right)\,\mathrm e^{-\xi^2}\mathrm e^{-i\frac{\omega}{a}\xi}\,\mathrm d\xi\\ &=\frac{1}{a}\mathrm e^{\frac{\omega^2}{4a^2}}\int_{-\infty}^\infty \text{erf}\left(\xi\right)\,\mathrm e^{-(\xi+i\frac{\omega}{2a})^2}\,\mathrm d\xi\\ &=\frac{1}{a}\mathrm e^{\frac{\omega^2}{4a^2}}\left[-\sqrt\pi\mathrm{erf}\left(i\frac{\omega}{2\sqrt 2\,a}\right)\right]\\ &=-\sqrt{8\pi\sigma}\mathrm e^{2\omega^2\sigma}\mathrm{erf}\left(i\omega\sqrt \sigma\right) \end{align} $$ using $$ \int_{-\infty}^\infty \text{erf}\left(x\right)\,\mathrm e^{-(\alpha x+\beta)^2}\,\mathrm dx=-\sqrt\pi\mathrm{erf}\left(\frac{\beta}{\sqrt{\alpha^2+1}}\right),\qquad\Re\{\alpha^2\}>-1 \tag{$\star$} $$

Using the imaginary error function $\mathrm{erfi}(z)=-i\,\mathrm{erf}(iz)$ we have

$$ F(\omega)=-i\sqrt{8\pi\sigma}\mathrm e^{2\omega^2\sigma}\mathrm{erfi}\left(\omega\sqrt \sigma\right) $$

Proof of $(\star)$

Let be $$F(\beta)=\int_{-\infty}^\infty \mathrm e^{-(\alpha x+\beta)^2}\text{erf}(x)\mathrm dx$$ and differentiating $$ \begin{align} F'(\beta)&=\frac{\mathrm d}{\mathrm d \beta} \left(\int_{-\infty}^\infty \mathrm e^{-(\alpha x+\beta)^2}\text{erf}(x)\mathrm dx \right)\\ &=-2\int_{-\infty}^\infty (\alpha x+\beta)e^{-(\alpha x+\beta)^2}\text{erf}(x)\mathrm dx \\&=\left[\mathrm e^{-(\alpha x+\beta)^2}\text{erf}(x) \right]_{-\infty}^\infty-\frac{2}{\sqrt{\pi}}\int_{-\infty}^\infty \mathrm e^{-(\alpha x+\beta)^2} \mathrm e^{-x^2}\mathrm dx\\ &=0-\frac{2}{\sqrt{\pi}}\sqrt{\frac{\pi}{\alpha^2+1}}\mathrm e^{-\frac{\beta^2}{\alpha^2+1}} \\ F'(\beta)&=-\frac{2}{\sqrt{\alpha^2+1}}\mathrm e^{-\frac{\beta^2}{\alpha^2+1}} \end{align} $$ and integrating this first order ODE we have $$ F(\beta)=F(0) -\frac{2}{\sqrt{\alpha^2+1}}\int_0^\beta\mathrm e^{-\frac{\xi^2}{\alpha^2+1}}\mathrm d\xi=F(0)-\sqrt\pi\text{erf}\left(\frac{\beta}{\sqrt{\alpha^2+1}}\right) $$ and observing that $F(0)=\int_0^0(\cdot)=0$ we finally have $$ \int_{-\infty}^\infty \mathrm e^{-(\alpha x+\beta)^2}\text{erf}(x)\mathrm dx=-\sqrt\pi\text{erf}\left(\frac{\beta}{\sqrt{\alpha^2+1}}\right) $$

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  • $\begingroup$ Thanks. But now, I would like to know how to show that conditional equality. I break down to half interval and integrate, which should have exponential function only! $\endgroup$ – will_cheuk Sep 13 '17 at 4:22
  • $\begingroup$ I'll add the details for the integral. $\endgroup$ – alexjo Sep 13 '17 at 10:36
  • $\begingroup$ But you didn't like the answer...did you? $\endgroup$ – alexjo Sep 19 '17 at 10:15
  • $\begingroup$ That's fine, I like it. But I think there should be a constant $\frac{1}{\alpha}$ missing in the final answer $\endgroup$ – will_cheuk Sep 22 '17 at 4:50

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