Fourier transform of wave packet

[schniefen]

Homework Statement

Consider $h(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \Re\{a(k)e^{i(kx-\omega t)}\}\mathrm{d}k.$ What is the Fourier transform $\hat{h}(k,t)$ evaluated at $t=0$, i.e. $\hat{h}(k,t=0)$? ($a(k)$ is given, but I do not think it needs to be specified)

Relevant Equations

The Fourier transform (FT) $\hat{f}(k,t)=\int_{-\infty}^{\infty} f(x,t)e^{-ikx}\mathrm{d}x$ and the inverse Fourier transform $f(x,t)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(k,t)e^{ikx}\mathrm{d}k$.

I am unsure if $h(x,t)$ really is a wave packet, but it looks like one, hence the title. Anyway, so I'd like to determine $\hat{h}(k,t=0)$. My attempt so far is recognizing that, without the real part in the integral, i.e.

$g(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} a(k)e^{i(kx-\omega t)} \mathrm{d}k,$
​

then $a(k)$ is just the Fourier transform of $g(x,t=0)$. However, I can not remove the real part from the integral and I am unsure how to proceed.

 

[DrClaude]

($a(k)$ is given, but I do not think it needs to be specified)

I disagree. What is a(k)?

 

[schniefen]

I disagree. What is a(k)?

$a(k)=A2\pi\delta(k)+2B/((k-C)^2+B^2)$, where $A, B$ and $C$ are real constants. $\delta$ is the Dirac delta.

 

[pasmith]

However, I can not remove the real part from the integral and I am unsure how to proceed.

For complex numbers $z$ and $w$, $$\Re(zw) = \Re(z)\Re(w) - \Im(z)\Im(w).$$ However for $w = e^{i\theta} = \cos \theta + i\sin \theta$ for real $\theta$ you may prefer to write $$\begin{split} \Re(e^{i\theta}) &= \frac{e^{i\theta} + e^{-i\theta}}2,\\ \Im(e^{i\theta}) &= \frac{e^{i\theta} - e^{-i\theta}}{2i}.\end{split}$$