# 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:

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.

• $f(x) = \frac{\pi}{2}e^{-x}\sqrt{x}\left(I_0(-x)-I_1(-x)\right)$ 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 $$h(2\pi n-s)+h(2\pi n+s)>0$$. $$\Box$$

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

• Thank you for this answer. Aug 2 '21 at 12:20