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The Fourier transform of $1$ is the (one-dimensional) Dirac delta function:

$$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^\infty dp\ e^{-i p x}. \tag{1}$$

Now I would like to replace the RHS with:

$$\frac{1}{2\pi} \int_{p_0}^\infty dp\ e^{-i p x}. \tag{2}$$

What happens to the LHS of $(1)$?

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2 Answers 2

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If we replace the RHS with $(2)$, we obtain the following:

$$\frac{1}{2\pi} \int_{p_0}^\infty dp\ e^{-ipx} = \frac{1}{2\pi} \int_{-\infty}^\infty dp\ \Theta(p - p_0) e^{-ipx} = \frac{1}{2\pi} e^{-i p_0 x} \int_{-\infty}^\infty dp\ \Theta(p) e^{-ipx}, \tag{3}$$

where the $\Theta$ function is the Heaviside step function. Then we just have to replace the LHS of $(1)$ with its Fourier transform, and we obtain:

$$\frac{1}{2\pi} \int_{p_0}^\infty dp\ e^{-ipx} = \frac{1}{2} e^{-i p_0 x} \left( \delta(x) - \frac{i}{\pi x} \right). \tag{4}$$

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I think Jxx has already gave one way of approaching this integral. I’ll give another, using a somewhat famous treatment in physics. So we wanna evaluate, $$I = \int_{p_0}^{\infty} dp e^{-ipx} = e^{-i p_0 x} \int_{0}^{\infty} dp e^{-ipx} = e^{-i p_0 x} \lim_{\Lambda \rightarrow \infty} \int_{0}^{\Lambda} dp e^{-ipx} = \lim_{\Lambda \rightarrow \infty} \frac{i e^{-i p_0 x} }{x}[e^{i\Lambda} -1].$$ The problem of course with this is that the limit is not defined. This actually shows up in many calculations in physics, the way to treat this is by introducing a small parameter $\delta$ to make the integration well defined then take the limit $\delta \rightarrow 0$ at the end. $$I = e^{-i p_0 x} \lim_{\delta \rightarrow 0} \int_{0}^{\infty} dp e^{-ip(x-i\delta)} = e^{-i p_0 x} \lim_{\delta \rightarrow 0} \frac{-i}{x-i\delta} .$$Note that I put $-i\delta$ rather than $i\delta$ to make the upper limit converge. Now one needs to be careful, you might think it’s possible to put $\delta = 0$ right away in the above expression, but it’s not. It’s a subtle point, but basically just doing so one loses information about the placement of the pole. In another words, imagine trying to integrate $\int_a^b dx\frac{1}{x-i\delta}$ such that $0\in(a,b)$, if you put $\delta = 0$ then you wouldn’t know how to evaluate this integral, in that sense putting $\delta =0$ in the above expression one loses information. Rather one way to proceed is the following, $$I = e^{-i p_0 x} \lim_{\delta \rightarrow 0} \frac{-i(x+i\delta)}{x^2+\delta^2} = e^{-i p_0 x} \lim_{\delta \rightarrow 0} \big[\frac{\delta}{x^2 + \delta^2} -\frac{ix}{x^2 + \delta^2} \big].$$ The first term here is in the limit of $\delta \rightarrow 0$ is $\pi \delta(x)$. Now for the second term one actually can actually put $\delta = 0$, under the assumption that if one where to integrate this function one takes the principal part because $\int_a^b dx \frac{x}{x^2+\delta^2} = P\int dx \frac{1}{x}$. So finally one may write, $$I = e^{-i p_0 x} \big[\pi \delta(x) - i P \frac{1}{x} \big] ,$$ where I left the $P$ there to remind ourselves that we take the principle part of that if we were to integrate this function.

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