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Consider the integral $\int_{-\infty}^\infty \frac{dz}{z - z_0}$. It has a simple pole at $z = z_0$. Assume $\Im (z_0) < 0$ so the pole is in lower half-plane. Divide

$$ \oint_{C_0} = \int_{-R}^R + \int_{C} $$

where we have enclosed the contour in the upper half-plane as $C = \{ R e^{i\theta} | \theta \in [0, \pi] \}$ is a large semicircle and $C_0 = C \cup [-R, R]$. Then

$$ \oint_{C_0} \frac{dz}{z - z_0} = 0. $$

Also

$$ \int_{C} \frac{dz}{z - z_0} = \begin{bmatrix} z = R e^{i\theta} \end{bmatrix} = \int_0^\pi \frac{R e^{i\theta}i d\theta}{R e^{i\theta} - z_0} = \int_0^\pi \frac{e^{i\theta}i d\theta}{e^{i\theta} - z_0/R}$$

Taking the limit $R \to \infty$,

$$ \int_{C} \frac{dz}{z - z_0} \to \int_0^\pi \frac{e^{i\theta}i d\theta}{e^{i\theta} - (0 + 0i)} = \int_0^\pi i d\theta = i\pi$$

so

$$\int_{-\infty}^\infty \frac{dz}{z - z_0} = - i \pi$$

Is this proof correct? Isn't the integral supposed to be undefined?

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    $\begingroup$ The integral doesn't exist as a Lebesgue or improper Riemann integral, but it exists as a principal value integral. You wrote $\Re (z_0) < 0$ where you meant $\Im (z_0) < 0$. With that fixed, you have a correct proof of the existence of the principal value. Another way, by translation you can assume $\Re (z_0) = 0$, and write $\frac{1}{x - iy} = \frac{x+iy}{x^2 + y^2}$. The integral of the real part has principal value $0$ since it is odd, the integral of the imaginary part exists as a Lebesgue integral. $\endgroup$ – Daniel Fischer May 19 '16 at 15:58
  • $\begingroup$ @DanielFischer Why doesn't it exist as improper Riemann integral? Isn't the integral in the complex plane along real axis always equal to Riemann integral? $\endgroup$ – Minethlos May 19 '16 at 18:25
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    $\begingroup$ $\int_{-\infty}^{\infty} f(x)\,dx$ exists as an improper Riemann integral if and only if ($f$ is Riemann integrable over all closed intervals $[a,b] \subset \mathbb{R}$ and) both $\lim\limits_{R\to +\infty} \int_0^R f(x)\,dx$ and $\lim\limits_{S\to +\infty} \int_{-S}^0 f(x)\,dx$ exist (in $\mathbb{C}$). Here, neither of the limits exist. You need some relation between $R$ and $S$ for $\lim \int_{-S}^R f(x)\,dx$ to exist. With $S = R$, you get the principal value $0$ for the real part. With $S = c\cdot R$, $c > 0$, you get the principal value $-\log c$ for the real part. $\endgroup$ – Daniel Fischer May 19 '16 at 18:39

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