Properties of laplace type transform of $t^{\alpha - 1}$ Let $p>2, \frac{1}{p} < \alpha < 1- \frac{1}{p}$ and define $g_\alpha(t) := t^{\alpha - 1} \chi_{[1, \infty]}$. Then $g_\alpha \in L^p(\mathbb R)$. Define $$f(z) := \int_1^\infty g_\alpha(t) \exp(-izt) \, dt.$$
for $z\in \mathbb C, \operatorname{Im }z <0.$ Then:  

$a)$ $f$ is holomorphic in the lower half plane,
$b)$ The limit $$f_0(x):= \lim_{\substack{y\to 0 \\ y < 0}} f(x+iy)$$ exists for all $x\neq 0$,
$c)$ $f(z)$ does not satisfy an estimate of the form $$\forall \epsilon >0: \quad \lvert f(z) \rvert \leq C_\epsilon \exp((1+\epsilon)\lvert z \rvert),$$
$d)$ $f_0$ is not $p$-integrable in any neighbourhood of $0$.

I managed to show $a)$ using Lebesgue's theorem on interchanging derivative and integral sign. However, I am unsure with the other assertions, especially $c)$ and $d)$: How does one estimate the integral from below? And how I can I show something about $f_0 $ if I don't explicitly know it? Any help appreciated!
 A: First we prove that
$$\tag{1}f(z) := \int_1^\infty t^{\alpha-1} e^{-izt} \, \mathrm{d} t$$
is convergent (as an improper Riemann integral) for all $z = x-iy$ with $x \ne 0$ and $y >0$. More presioulsy, the convergence is uniform if $|x| \ge x_0$ for fixed $x_0>0$. Thus (1) defines a continuous function and we have for $x \ne 0$ that
$$\tag{2}f_0(x) = \int_1^\infty t^{\alpha-1} e^{-ixt} \, \mathrm{d} t$$
Prove: For any $z= x-iy$ with $|x| \ge x_0$ and $y \ne 0$ we have with $1 \le a < b$ that
\begin{align}
\tag{3}\int_a^b t^{\alpha-1} e^{-izt} dt = \frac{1}{iz} ( a^{\alpha-1} e^{-iza} -  b^{\alpha-1} e^{-izb}) - \frac{\alpha-1}{iz} \int_a^b t^{\alpha-2} e^{-izt} dt.
\end{align}
The last line can be bounded by
$$\frac{1}{x_0} (a^{\alpha-1} + b^{\alpha-1}) +\frac{1-\alpha}{x_0} \int_a^b t^{\alpha-2} dt \le \frac{2}{x_0} a^{\alpha-1} $$
and thus (3) is a Cauchy sequence and thus convergent. In fact, it is (uniformly) convergent and thus continuous. (The whole argument is also known as Dirchlet's test, see for example here.)
Prove of (c) and (d): So we have shown (b). With the help of the explicit identity (2) we can also verify (c) and (d). In fact, we have (after coordinate of change) the identity
$$f_0(x) = x^{-\alpha} \int_x^\infty s^{\alpha-1} e^{-is} \, \mathrm{d} s.$$
Since the integrand is locally integrable (also in $s=0$) the function
$$g(x) := \int_x^\infty s^{\alpha-1} e^{-is} \, \mathrm{d} s$$
is continuous in $x=0$ with
$$\tag{4}g(0) = \int_0^\infty s^{\alpha-1} e^{-is} \, \mathrm{d} s.$$
Thus, it remains to show that $g(0) \ne 0$. Here we use the integral representation of (4). In fact, (4) is related to the Gamma function. One representation of the Gamma function (which was proven by Euler by using a contour shift argument) is
$$\Gamma(w) = e^{i \pi w /2} \int_0^\infty s^{w-1} e^{-is} \, \mathrm{d} s$$
if $0 < \mathrm{Re}(w) < 1$. Hence
$$g(0) = e^{-i \alpha \pi/2} \Gamma(\alpha) \neq 0.$$
All in all, we see that
$$\tag{5} f_0(x) \sim_\alpha x^{-\alpha} \quad (\text{for} \ x \rightarrow 0) $$
up to an non-zero constant (depending on $\alpha$).
c) can not hold, because c) would imply that $f_0$ is bounded. (5) implies that $f_0$ is not $p$-integrable (note that $\alpha p >1$) in any neighbourhood of $0$.
