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... 

(Here is a link to the page on Google Books.)
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?)
 A: 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
$$\int_0^\infty e^{-ixt} \, dt = -i \mathrm{P}\frac{1}{x} + \pi \, \delta(x)$$
