Limit $\lim _{t\to \infty} \int_t^\infty t \frac {\sin^2 (u)}{u^2} du$

Question

Does the limit $\displaystyle{\lim _{t \to \infty} t \cdot \int_t^\infty \frac {\sin^2 (u)}{u^2} \mathrm{d}u}$ exist?

Intuitively, since $\sin(t) > \frac 1 2$ holds for at least half of a period, I would assume that the integral $\int_t^\infty \frac {\sin^2 (u)}{u^2} \mathrm{d}u$ behaves approximately as $\int_t^\infty \frac {1}{u^2} \mathrm{d}u$. But the latter is $\frac 1 t$, so it seems that the limit diverges. Is it true?

Answer 1

$$I(t)= t \cdot \int_t^\infty \frac {\sin^2 (u)}{u^2} \mathrm{d}u$$

Let $x=u/t$

$$I(t)=\int_1^\infty \frac{\sin^2 tx}{x^2}dx$$

Integrate by parts and define $y=2tx$, to get

$$I(t)=\sin^2 t+t\int_{2t}^\infty \frac{\sin y}{y}dy$$

Integration by part again,

$$I(t)=\sin^2 t+\frac{1}2\cos2t-t\int_{2t}^\infty \frac{\cos y}{y^2}dy=\frac{1}2-t\int_{2t}^\infty \frac{\cos y}{y^2}dy$$

Next, we show the limit of the last term is $0$

$$\lim_{t\rightarrow \infty} t\int_{2t}^\infty \frac{\cos y}{y^2}dy=\lim_{t\rightarrow \infty} t\cdot\frac{\sin 2t}{4t^2}+2t\int_{2t}^\infty \frac{\sin y}{y^3}dy=\lim_{t\rightarrow \infty} 2t\int_{2t}^\infty \frac{\sin y}{y^3}dy$$ and

$$\lim_{t\rightarrow \infty} \left|2t\int_{2t}^\infty \frac{\sin y}{y^3}dy\right|\le \lim_{t\rightarrow \infty} \left|2t\int_{2t}^\infty \frac{1}{y^3}dy\right|=0$$

Therefore,

$$\lim_{t\rightarrow \infty}I(t)=\frac{1}2$$

Answer 2

Unfortunately you can not use calculate it using de L'Hopital and Leibniz integral Rule: $$\lim_{x\to \infty}x\cdot\int_{x}^{\infty}\frac{\sin(t)^2}{t^2}\mathrm{d}t\\ \lim_{x\to \infty}\dfrac{\int_{x}^{\infty}\frac{\sin(t)^2}{t^2}\mathrm{d}t}{\frac{1}{x}}\overset{H}{=}\lim_{x\to \infty}\dfrac{\frac{\mathrm{d}}{\mathrm{d}x}\int_{x}^{\infty}\frac{\sin(t)^2}{t^2}\mathrm{d}t}{\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{x}}\\ \lim_{x\to \infty}\dfrac{\frac{\mathrm{d}}{\mathrm{d}x}\int_{x}^{\infty}\frac{\sin(t)^2}{t^2}\mathrm{d}t}{-\frac{1}{x^2}}$$ Leibniz integral rule is $${\displaystyle {\frac {d}{dx}}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)} =f{\big (}x,b(x){\big )} {\frac {d}{dx}}b(x)-f{\big (}x,a(x){\big )} {\frac {d}{dx}}a(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}f(x,t)\,dt$$ So you have $$\frac{d}{dx}\int_x^{\infty}\frac{\sin(t)^2}{t^2}\mathrm{d}t=\lim_{x\to\infty}\frac{\sin(x)^2}{x^2}\cdot \underbrace{0}_{\frac{d}{dx}\infty}-\frac{\sin(x)^2}{x^2}\cdot \underbrace{1}_{\frac{d}{dx} x}+\int_{x}^{\infty}\underbrace{0}_{\frac{d}{dx}\frac{\sin(t)}{t^2}}\mathrm{d}t$$ The final limits it would turn out$$\lim_{x\to\infty}\dfrac{-\frac{\sin(x)^2}{x^2}}{-\frac{1}{x^2}}=\sin(x)^2$$ But this is undeterminate.
You can try calculating the integral of the function: $$\int_{x}^{\infty}\frac{\sin(t)^2}{t^2}\mathrm{d}t=\frac{\pi}{2}-\text{Si}(2 x) + \frac{\sin(x)^2}{x}$$ Where $\displaystyle\text{Si}(z):=\int_{0}^{z}\dfrac{\sin(t)}{t}\mathrm{d}t$ is the integral sine function.
Is known that $\text{Si}(2x)\sim\dfrac{\pi}{2}-\dfrac{\sin(2x)}{4x^2} - \dfrac{\cos(2 x)}{2x}$
So the final limit is $$\lim_{x\to\infty}x\left(\frac{\pi}{2}-\text{Si}(2 x) + \frac{\sin(x)^2}{x}\right)=\ \lim_{x\to\infty}x\left(\frac{\pi}{2}-\dfrac{\pi}{2}+\dfrac{\sin(2x)}{4x^2} +\dfrac{\cos(2 x)}{2x} + \frac{\sin(x)^2}{x}\right)=\\ \lim_{x\to\infty}x\left(\dfrac{\sin(2x)}{4x^2} +\dfrac{\cos(2 x)}{2x} + \frac{\sin(x)^2}{x}\right)=\\ \lim_{x\to\infty}x\left(\dfrac{\sin(2x)}{4x^2} +\dfrac{\cos(2 x)}{2x} + \frac{\sin(x)^2}{x}\right)=\\ \lim_{x\to\infty}\underbrace{\frac{\sin(2x)}{4x}}_{\to 0}+\underbrace{\frac{\cos(2x)}{2}+\sin(x)^{2}}_{=\frac{1}{2}}=\frac{1}{2} $$

Answer 3

Change of variable $x=u/t$ yields

$$t\int^\infty_t\frac{\sin^2u}{u^2}\,du=t\int^\infty_1\frac{\sin^2(tx)}{t^2x^2}\,tdx=\int^\infty_1\frac{\sin^2 (tx)}{x^2}dx$$

By Fejer's formula $$ \int^\infty_1\frac{\sin^2 (tx)}{x^2}dx\xrightarrow{t\rightarrow\infty}\Big(\frac{1}{\pi}\int^\pi_0\sin^2x\,dx\Big)\int^\infty_1\frac{dx}{x^2}=\frac12 $$

Answer 4

\begin{align*} &t\int_{t}^{\infty}\frac{\sin^{2}u}{u^{2}}du-\frac{1}{2}\\ &=t\int_{t}^{\infty}\frac{\sin^{2}u}{u^{2}}du-\frac{t}{2}\int_{t}^{\infty}\frac{1}{u^{2}}du\\ &=t\int_{t}^{\infty}\frac{2\sin^{2}u-1}{2u^{2}}du\\ &=t\int_{t}^{\infty}\frac{-\cos 2u}{2u^{2}}du\\ &=-\frac{t}{2}\left[\frac{\sin 2u}{2}\cdot\frac{1}{u^{2}}\bigg|_{u=t}^{u=\infty}-\frac{1}{2}\int_{t}^{\infty}\frac{-2\sin 2u}{u^{3}}du\right]\\ &=-\frac{t}{2}\left[-\frac{\sin 2t}{2t^{2}}+2\int_{t}^{\infty}\frac{\sin 2u}{u^{3}}du\right]. \end{align*} Now we have \begin{align*} t\int_{t}^{\infty}\Big|\frac{\sin 2u}{u^{3}}\Big|du\leq t\int_{t}^{\infty}\frac{1}{u^{3}}du=t\cdot\frac{1}{2}\cdot\frac{1}{t^{2}}\rightarrow 0. \end{align*}