I'm reading about asymptotic analysis on the integral $I(x)=\int_0^1{\ln{t}e^{ixt}}dt$. The book tells me that I can use Cauchy theorem to deform the contour into a rectangular contour: $0 \to iT \to 1+iT \to 1 \to 0$ (where $T$ is any finite number).
My question is:
Isn't $\ln{t}e^{ixt}$ singular at $t=0$? Why can I apply Cauchy theorem then?
Do I add a quarter circle around the origin to avoid the singularity? I don't think it works because if you let $z=re^{i\theta}$, then $\lim_{r\to 0}{\int_0^{\pi /2}{\ln{(z)}e^{ixz}}d\theta}\neq 0$.
We could show what you want using the estimation lemma.
Let $C_r$ be the quarter circle of radius $r$ that you're interested in. Then, since $$|\log(z)| = \sqrt{(\log |z|)^2 + (\arg z)^2},$$ we have
$$ \begin{align} \left|\int_{C_r} \log(z) e^{ixz}\,dz\right| &\leq \frac{\pi}{2} r \cdot \max_{z \in C_r} \left|\log(z) e^{ixz}\right| \\ &\leq \frac{\pi}{2} r \sqrt{(\log r)^2 + (\pi/2)^2} \cdot \max_{z \in C_r} \left|e^{ixz}\right| \\ &= \frac{\pi}{2} r \sqrt{(\log r)^2 + (\pi/2)^2} \cdot \max_{\theta \in [0,\pi/2]} e^{-xr \sin \theta} \\ &= \frac{\pi}{2} r \sqrt{(\log r)^2 + (\pi/2)^2} \\ &= \frac{\pi}{2} r |\log r| \sqrt{1 + \pi^2/(2 \log r)^2} \end{align} $$
for $0 < r < 1$. Then, since $r \log r \to 0$ as $r \to 0^+$, the result follows.
Alternately, we could show it by parameterizing the contour.
Letting $z = r e^{i\theta}$ yields
$$ \int_{C_r} \log(z) e^{ixz}\,dz = i r \int_0^{\pi/2} \log(r e^{i \theta}) e^{i x r e^{i\theta}} e^{i \theta}\,d\theta. $$
Now
$$ \log(r e^{i\theta}) = \log r + i\theta, $$
so we can rewrite this as
$$ i r \log r \int_0^{\pi/2} e^{i x r e^{i\theta}} e^{i\theta}\,d\theta - r \int_0^{\pi/2} \theta e^{i x r e^{i\theta}} e^{i\theta}\,d\theta. $$
Since
$$ \int_0^{\pi/2} e^{i x r e^{i\theta}} e^{i\theta}\,d\theta \longrightarrow \int_0^{\pi/2} e^{i\theta}\,d\theta = 1+i, $$
$$ \int_0^{\pi/2} \theta e^{i x r e^{i\theta}} e^{i\theta}\,d\theta \longrightarrow \int_0^{\pi/2} \theta e^{i\theta}\,d\theta = \frac{\pi}{2} - 1 + i, $$
and
$$ r\log r \longrightarrow 0 $$
as $r \to 0^+$, we see that
$$ \int_{C_r} \log(z) e^{ixz}\,dz \longrightarrow 0 $$
as $r \to 0^+$.