Monotonicity of a parametric integral For real $x>0$, let
$$f(x):=\frac1{\sqrt x}\,\int_0^\infty\frac{1-\exp\{-x\, (1-\cos t)\}}{t^2}\,dt.$$

How to prove that $f$ is increasing on $(0,\infty)$?

Here is the graph $\{(x,f(x))\colon0<x<3\}$:

So, it even appears that $f$ is concave. Since $f>0$, the concavity would of course imply that $f$ is increasing.
This question arises in a Fourier analysis of a certain probabilistic problem.
 A: 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 $h(2\pi n-s)+h(2\pi n+s)>0$. $\Box$
A: Here is another solution: Noting the identity
$$\sum_{n=-\infty}^{\infty} \frac{1}{(t+2\pi n)^2} = \frac{1}{2(1-\cos t)}$$
and taking advantage of the fact that the integrand is non-negative, we may apply Tonelli's theorem to get
\begin{align*}
f(x)
&= \frac{1}{2\sqrt{x}} \int_{-\pi}^{\pi} \left( \sum_{n=-\infty}^{\infty} \frac{1}{(t+2\pi n)^2} \right) \bigl( 1 - e^{-x\left(1-\cos t\right)} \bigr) \, \mathrm{d}t \\
&= \frac{1}{4\sqrt{x}} \int_{-\pi}^{\pi} \frac{1 - e^{-x\left(1-\cos t\right)}}{1-\cos t} \, \mathrm{d}t \\
&= \frac{1}{8\sqrt{x}} \int_{-\pi}^{\pi} \frac{1 - e^{-2x\sin^2(t/2)}}{\sin^2(t/2)} \, \mathrm{d}t. \tag{1}
\end{align*}
Then by applying the integration by parts, we also get
$$ f(x) = \frac{\sqrt{x}}{2} \int_{-\pi}^{\pi} \cos^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t. \tag{2} $$
Now using $\text{(1)}$,
$$ \bigl( \sqrt{x}f(x) \bigr)' = \frac{1}{4} \int_{-\pi}^{\pi} e^{-2x\sin^2(t/2)} \, \mathrm{d}t. \tag{3} $$
Therefore by $\text{(2)}$ and $\text{(3)}$ altogether,
\begin{align*}
\sqrt{x}f'(x)
&= \bigl( \sqrt{x}f(x) \bigr)' - \frac{1}{2\sqrt{x}} f(x) \\
&= \frac{1}{4} \int_{-\pi}^{\pi} e^{-2x\sin^2(t/2)} \, \mathrm{d}t -\frac{1}{4} \int_{-\pi}^{\pi} \cos^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t \\
&= \frac{1}{4} \int_{-\pi}^{\pi} \sin^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t \\
&> 0.
\end{align*}
This identity can also be used to show that $f''(x) < 0$, and hence $f$ is concave as expected.
