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.

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.
