Fourier Transform of a Wave Packet [closed]

Question

In the analysis of coherence and interference, I encountered the following expression: $$F(t)=\Re\int_0^\infty\mathrm d\omega e^{-i\omega t}H(\omega)$$, where $\Re$ denotes the real part of the integral, and $H(\omega)$ vanishes for $\omega<0$. I want to show the Fourier transform of $F(t)$ is $$\hat{F}(t)=\frac{\sqrt {2\pi}}{2}[H(\omega)+H^\star(-\omega)]$$ Here is my approach $$\int_0^\infty\mathrm d\omega e^{-i\omega t}H(\omega)=\int_{-\infty}^\infty\mathrm d\omega e^{-i\omega t}H(\omega)=\sqrt{2\pi}H(t)$$ where $H(t)$ is the Fourier transform of $H(\omega)$ (with out the hat), and I used the fact $H(\omega)=0$ for $\omega<0$. Hence $$F(t)=\frac{\sqrt {2\pi}}{2}[H(t)+H^\star(t)]$$ It follows that $$\hat F(t)=\frac{\sqrt {2\pi}}{2}\int_{-\infty}^\infty \mathrm d\omega e^{-i\omega t}[H(t)+H^\star(t)]=\pi[H(\omega)+H^\star(-\omega)]$$ Although the form of my result agree with the correct answer, I got a factor of $\sqrt{2\pi}$ wrong somewhere. I wonder where did I miss it!

Answer 1

Mathematicians usually put $1/\sqrt{2\pi}$ before each integral. Physicists usually leave it out of the integral like your first equation, but then divide the inverse integral by $2\pi$