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Matt L.
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For $z$ outside $C$ we have, it follows from Cauchy's theorem that

For $z$ outside $C$ we have

For $z$ outside $C$, it follows from Cauchy's theorem that

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Matt L.
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Depending on the field, the Hilbert transform relations $(11)$ and $(12)$ are also referred to as Kramers-Kronig relations or as dispersion relations (e.g., here and here).

Example 3:

Example 3:

Depending on the field, the Hilbert transform relations $(11)$ and $(12)$ are also referred to as Kramers-Kronig relations or as dispersion relations (e.g., here and here).

Example 3:

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Matt L.
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Example 3:

Given the causal and stable impulse response

$$h(t)=e^{-t}u(t)$$

we know from the above that the corresponding transfer function

$$H(s)=\frac{1}{1+s}$$

must be analytic and decaying in the right half-plane, both of which is straightforward to verify. Furthermore, $H(s)$ must satisfy the Hilbert transform relations $(11)$ and $(12)$ on the imaginary axis. Splitting $H(j\omega)$ into real and imaginary parts gives

$$\mathscr{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}$$

which is a well-known Hilbert transform pair (see this answer or this table).

Example 3:

Given the causal and stable impulse response

$$h(t)=e^{-t}u(t)$$

we know from the above that the corresponding transfer function

$$H(s)=\frac{1}{1+s}$$

must be analytic and decaying in the right half-plane, both of which is straightforward to verify. Furthermore, $H(s)$ must satisfy the Hilbert transform relations $(11)$ and $(12)$ on the imaginary axis. Splitting $H(j\omega)$ into real and imaginary parts gives

$$\mathscr{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}$$

which is a well-known Hilbert transform pair (see this answer or this table).

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