5
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ $f(x) = \frac{\pi}{2}e^{-x}\sqrt{x}\left(I_0(-x)-I_1(-x)\right)$ $\endgroup$ Jul 1 at 2:06
3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Is this answer complete (and correct)? If so, can you mark the check? $\endgroup$ Jul 11 at 16:48
  • 1
    $\begingroup$ @mathworker21 : I have now marked the answer. $\endgroup$ Jul 11 at 18:04
2
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for this answer. $\endgroup$ Aug 2 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.