Prinicipal value integral of $\exp(iaz)/z$ I'm supposed to show that $$p.v. \int_{\mathbb{R}}\frac{e^{iat}}{t}dt = i\pi \operatorname{sgn}(a)$$
by integrating along
the two semi-circles.
So far I have the following for a>0:
\begin{align}
p.v. \int_{\mathbb{R}}\frac{e^{iat}}{t}dt&=\lim_{r\to 0,R\to\infty}\int_{r<|t|<R}\frac{e^{iat}}{t}dt
\\&=\lim_{r\to 0,R\to\infty}\int_{\gamma_{-}}\frac{e^{iat}}{t}dt+\int_{\gamma{+}}\frac{e^{iat}}{t}dt\stackrel{Cauchy}{=}\lim_{r\to 0}\int_{\gamma_r}\frac{e^{iat}}{t}dt-\lim_{R\to\infty}\int_{\gamma_R}\frac{e^{iat}}{t}dt
\\&=\lim_{r\to0}i\int_{0}^{\pi}e^{iare^{i\phi}}d\phi-\lim_{R\to\infty}i\int_{0}^{\pi}e^{iaRe^{i\phi}}d\phi
\\&=i\pi-\lim_{R\to\infty}i\int_0^\pi \underbrace{e^{iaR\cos\phi}}_{\text{bound by }1}\cdot \underbrace{e^{-aR\sin\phi}}_{\to 0}=i\pi
\end{align}

*

*I'm not sure if I'm allowed to do that last step to show that the second integral is $0$.

*Also I'm struggling with $a<0$ because if that last step would be correct the second integral would go to infinity. Or do I have to calculate that completely different?

*Can I use the normal logarithm to show that the integral is $0$ for $a=0$?


Any help would be very much appreciated :)

 A: Question 1
No, this is not allowed. Indeed, if $a>0$ and if $\phi\in(0,\pi)$ the factor $\lim_{R\to\infty}\exp(-aR\sin\phi)=0$. But note that we have to include the endpoints $0$ and $\pi$, which means that for them $\lim_{R\to\infty}\exp(-aR\sin\phi)=1$.
To prove that the absolut value of the integral is bounded by zero, you can use Jordans inequality. To be more specific, you can use Jordans Lemma, which uses Jordans inequality in its proof.
Question 2
If $a$ is negative, the inequality is still correct but we cannot deduce the value of the integral. This means that the absolut value of the integral over the semicircle is bounded by infinity, which includes every non-negative real number.
To avoid this, you have to integrate over the semicircle in the lower half of the complex plane. In this case Jordans Lemma is also applicable.
Question 3
Yes, you can use the "normal" logarithm to show that the integral is equal to zero. By the definition of the principal value,
$$\operatorname{PV} \int_{\mathbb{R}}\frac{1}{t}\,\mathrm{d}t=\lim_{\varepsilon\to0^+}\left(\int_{-\infty}^{-\varepsilon}\frac{1}{t}\,\mathrm{d}t+\int_{\varepsilon}^{\infty}\frac{1}{t}\,\mathrm{d}t\right)$$
and using logarithm rules after integrating we get
$$\lim_{\varepsilon\to0^+}\log(-\varepsilon)-\lim_{R\to\infty}\log( -R)+\lim_{\varepsilon\to0^+}\log(\varepsilon)-\lim_{R\to\infty}\log(R)$$
and hence
$$\lim_{\varepsilon\to0^+}\log(-\frac{\varepsilon}{\varepsilon})-\lim_{R\to\infty}\log(-\frac{R}{R})=i-i=0.$$
Thus the integral satisfies the general formula $i\pi\operatorname{sgn}(a)$. There is also a faster way for $a=0$. If we substitute $t\to-t$ in the first equation, where we used the definition of the principal value, the two integrals are equal up to a sign, which then eliminate each other.
