Show that $\int_0^1 |t-z|^{-1/2}\ \mathrm{d}t < c(1 + |z|)^{-1/2}$ I want to show that there is a constant $c > 0$ such that
$$
\int_0^1 |t-z|^{-1/2}\ \mathrm{d}t < c(1 + |z|)^{-1/2}
$$
for any $z \in \mathbb{C}$.
I found the assertion in a paper I'm reading and I have not succeeded in proving it. Perhaps it uses a common technique in complex analysis (my complex analysis is rusty). Power series expansions came to mind, but I still don't see how to make use of it.
 A: This follows from straight up estimates:
If $|z|>2$ then $|1-t/z| >  1/2$ for $0 \le t \le 1$ so $$\int_0^1 |t-z|^{-1/2}\ \mathrm{d}t = |z|^{-1/2}\int_0^1 |t/z-1|^{-1/2}\ \mathrm{d}t < \sqrt 2|z|^{-1/2}< 2(1+|z|)^{-1/2}$$ since $(1+|z|)^{1/2}<(2|z|)^{1/2}, |z| >2$
If $|z| \le 2$ the function $f(z)=\int_0^1 |t-z|^{-1/2}\ \mathrm{d}t$ is continuous on the disc $|z| \le 2$ (if $u \in [0,1]$ one can apply the dominated convergence theorem to show $f(z_n) \to f(u), z_n \to u$, while for $z$ not in $[0,1]$, the result is straightforward as the integrand is continuous in a small neighborhood of $z$ then), so $|f(z)| \le C, |z| \le 2$ hence $|f(z)| < 2C(1+|z|)^{-1/2}, |z| \le 2$ since $(1+|z|)^{-1/2} \ge 3^{-1/2}$
Putting the above together one gets $|f(z)| < \max (2,2C)(1+|z|)^{-1/2}$ for all complex $z$
A: Alternative proof:
Let $z = a + b\mathrm{i}$.
We have
$$(1 + |z|)^{-1/2} = \frac{1}{\sqrt{1 + \sqrt{a^2 + b^2}}} \ge \frac{1}{\sqrt{1 + |a| + |b|}}. \tag{1}$$
Also, we have
$$|t - z| = \sqrt{(t - a)^2 + b^2} \ge \frac{|t - a| + |b|}{\sqrt2}$$
and thus
$$\int_0^1 |t - z|^{-1/2}\mathrm{d} t
\le \int_0^1 \frac{\sqrt[4]{2}}{\sqrt{|t - a| + |b|}}\mathrm{d}t =: I.$$
If $a\le 0$, we have
$$I = \int_0^1 \frac{\sqrt[4]{2}}{\sqrt{t - a + |b|}}\mathrm{d}t
= \frac{2\sqrt[4]{2}}{\sqrt{|a| + |b| + 1} + \sqrt{|a| + |b|}}
\le \frac{2\sqrt[4]{2}}{\sqrt{|a| + |b| + 1}}. \tag{2}$$
If $a \ge 1$, we have
\begin{align*}
 I &= \int_0^1 \frac{\sqrt[4]{2}}{\sqrt{a - t + |b|}}\mathrm{d}t\\
  &= \frac{2\sqrt[4]{2}}{\sqrt{|a| + |b|} + \sqrt{|a| + |b| - 1}}\\
  &\le \frac{2\sqrt[4]{2}}{\sqrt{|a| + |b|}}\\
  &\le \frac{4\sqrt[4]{2}}{\sqrt{|a| + |b| + 1}}.\tag{3}
\end{align*}
If $0 < a < 1$, we have
\begin{align*}
 I &= \int_0^a \frac{\sqrt[4]{2}}{\sqrt{a - t + |b|}}\mathrm{d}t
 + \int_a^1 \frac{\sqrt[4]{2}}{\sqrt{t - a + |b|}}\mathrm{d}t\\
 &= \frac{2a\sqrt[4]{2}}{\sqrt{a + |b|} + \sqrt{|b|}}
 + \frac{2(1 - a)\sqrt[4]{2}}{\sqrt{1 - a + |b|} + \sqrt{|b|}}\\
 &\le \frac{2a\sqrt[4]{2}}{\sqrt{a + |b|}}
 + \frac{2(1 - a)\sqrt[4]{2}}{\sqrt{1 - a + |b|}}\\
 &\le \frac{4\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}
 + \frac{4\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}\\
 &= \frac{8\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}.\tag{4}
\end{align*}
Remark: Here it is easy to prove that
$\frac{2a\sqrt[4]{2}}{\sqrt{a + |b|}}
\le \frac{4\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}$
and
$\frac{2(1 - a)\sqrt[4]{2}}{\sqrt{1 - a + |b|}}
\le \frac{4\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}$.
Using (1)-(4), we have
$$I < \frac{16}{\sqrt{1 + |a| + |b|}} \le 16(1 + |z|)^{-1/2}.$$
We are done.
