We have to show that the function $g$ defined by
\begin{equation*}
g(y):= f(y^2):=\frac1y\,\int_0^\infty\frac{1-\exp\{-y^2\, (1-\cos t)\}}{t^2}\,dt
\end{equation*}
is increasing on $(0,\infty)$.

Note that
\begin{equation*}
g'(y)=-\frac1{y^2}\,I_1+2I_2,
\end{equation*}
where
\begin{equation*}
\begin{aligned}
I_1&:=\int_0^\infty\frac{1-\exp\{-y^2\, (1-\cos t)\}}{t^2}\,dt, \\
I_2&:=\int_0^\infty2\frac{1-\cos t}{t^2}\,\exp\{-y^2\, (1-\cos t)\}\,dt.
\end{aligned}
\end{equation*}
Taking integral $I_1$ by parts, we get
\begin{equation*}
g'(y)=\int_0^\infty h(t)\,dt=j_0+\sum_{n=1}^\infty J_n, \tag{1}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
h&:=h_1h_2,\\
h_1(t)&:=2\frac{1-\cos t}{t^2}-\frac{\sin t}t, \\
h_2(t)&:=\exp\{-y^2\, (1-\cos t)\}, \\
j_0&:=\int_0^\pi h(t)\,dt, \\
J_n&:=\int_0^\pi [h(2\pi n-s)+h(2\pi n+s)]\,ds.
\end{aligned}
\tag{2}
\end{equation*}

**Lemma 1:** $h>0$ on $(0,2\pi)$.

**Lemma 2:** $h(2\pi n-s)+h(2\pi n+s)>0$ for all natural $n$ and all $s\in(0,\pi)$.

It follows from (1), (2), and Lemmas 1, 2 that $g'>0$ on $(0,\infty)$ and hence $g$ is indeed increasing on $(0,\infty)$.

It remains to prove Lemmas 1, 2.

*Proof of Lemma 1:* Let $H(t):=t^2 h_1(t)$. Then $H(0)=H'(0)=0=H(2\pi)$ and $H''(t)=t\sin t$, so that $H$ is strictly convex on $[0,\pi]$ and strictly concave on $[\pi,2\pi]$. So, $H>0$ and hence $h_1>0$ on $(0,2\pi)$. Also, $h_2>0$ everywhere. Now Lemma 1 follows. $\Box$

*Proof of Lemma 2:* For any natural $n$ and any $s\in(0,\pi)$, we have $h_2(2\pi n-s)=h_2(2\pi n+s)>0$ and
\begin{equation*}
h_1(2\pi n-s)+h_1(2\pi n+s)
=2\frac{1-\cos s}{(2\pi n-s)^2}+2\frac{1-\cos s}{(2\pi n+s)^2}
+\frac{\sin s}{2\pi n-s}-\frac{\sin s}{2\pi n+s},
\end{equation*}
which is manifestly $>0$ and hence $h(2\pi n-s)+h(2\pi n+s)>0$. $\Box$