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 $\{(x,f(x))\colon0<x<3\}$:

enter image description here

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.

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    $\begingroup$ $f(x) = \frac{\pi}{2}e^{-x}\sqrt{x}\left(I_0(-x)-I_1(-x)\right)$ $\endgroup$ Jul 1 '21 at 2:06

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 $h(2\pi n-s)+h(2\pi n+s)>0$. $\Box$

  • $\begingroup$ Is this answer complete (and correct)? If so, can you mark the check? $\endgroup$ Jul 11 '21 at 16:48
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    $\begingroup$ @mathworker21 : I have now marked the answer. $\endgroup$ Jul 11 '21 at 18:04

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.

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    $\begingroup$ Thank you for this answer. $\endgroup$ Aug 2 '21 at 12:20

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