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- 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.