# Half-range Fourier transform of $e^{-ixt}$

I'm reading a textbook that has the following sentence:

Making use of the formula $$\int_0^\infty dt e^{-i x t} = \pi \delta (x) - i \text{P}\frac{1}{x} ,$$ where $\text{P}$ denotes the Cauchy principal value...

Using the Fourier transform $\int_{-\infty}^{\infty}dt e^{-ixt} = 2\pi\delta(x)$, I can show that the real part of the integral is $$\text{Re}\int_0^\infty dt e^{-i x t} = \pi \delta (x) .$$ But how can I show that the imaginary part is $$\text{Im}\int_0^\infty dt e^{-i x t} = -i\text{P}\frac{1}{x} ,$$ and what does the Cauchy principal value mean in this context? (i.e., principal value of what?)

• $\mathrm{P}\frac{1}{x}$ is a distribution defined by $$\left< \mathrm{P}\frac{1}{x}, \phi(x) \right> = \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac{1}{x} \phi(x) \, dx$$ – md2perpe Sep 25 '17 at 20:11

First note that $$\int_0^\infty e^{-ixt} \, dt = \int_{-\infty}^{\infty} H(t) \, e^{-ixt} \, dt$$ where $H$ is the Heaviside step function.

Now we multiply the above identity with $x$: $$x \int_0^\infty e^{-ixt} \, dt = x \int_{-\infty}^{\infty} H(t) \, e^{-ixt} \, dt = \int_{-\infty}^{\infty} H(t) \, x e^{-ixt} \, dt = \int_{-\infty}^{\infty} H(t) \, i \frac{d}{dt} \left(e^{-ixt}\right) \, dt \\ = \{ \text{ partial integration } \} = -i \int_{-\infty}^{\infty} H'(t) \, e^{-ixt} \, dt = -i \int_{-\infty}^{\infty} \delta(t) \, e^{-ixt} \, dt = -i$$

The solutions to the distribution equation $x \, u(x) = 1$ are $u(x) = \mathrm{P} \frac{1}{x} + C \delta(x)$ where $C$ is some constant. Therefore we get $$\int_0^\infty e^{-ixt} \, dt = -i \mathrm{P}\frac{1}{x} + C \, \delta(x)$$

By symmetry reasoning we find that $$\int_{-\infty}^0 e^{-ixt} \, dt = i \mathrm{P}\frac{1}{x} + C \, \delta(x)$$

Thus $$\int_{-\infty}^{\infty} e^{-ixt} \, dt = 2C \, \delta(x)$$ but it's well-known that $$\int_{-\infty}^{\infty} e^{-ixt} \, dt = 2\pi \, \delta(x)$$

Hence we conclude that $C = \pi$ which gives $$\int_0^\infty e^{-ixt} \, dt = -i \mathrm{P}\frac{1}{x} + \pi \, \delta(x)$$

• Thank you. I'm a bit stuck on why $u(x) = \text{P}\frac{1}{x} + C\delta(x)$ is the solution to $xu(x) =1$, though. Could you tell me the reasoning behind that (or point me to a reference)? – user37921 Sep 26 '17 at 9:31
• There's some information on Wikipedia. – md2perpe Sep 26 '17 at 16:17
• It's easy to show that $x \mathrm P \frac1x = 1:$ $$\langle x \mathrm P \frac1x, \phi(x) \rangle = \langle \mathrm P \frac1x, x \phi(x) \rangle = \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac1x \cdot x \phi(x) \, dx \\ = \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \phi(x) \, dx = \int \phi(x) \, dx = \langle 1, \phi \rangle$$ for all $\phi \in C_c^\infty(\mathbb R).$ – md2perpe Sep 26 '17 at 16:26
• Then, if $v$ is another distribution satisfying $x \, v(x) = 1$ then we have $x \, (u-v) = 0.$ The only distributions satisfying $x \, w(x) = 0$ are $w(x) = C \, \delta(x)$ so we must have $u-v = C \, \delta.$ – md2perpe Sep 26 '17 at 16:29