Show that $\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t}{t^2}dt=\ln 2.$

Question

Question: Show that $$\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t}{t^2}dt=\ln 2.$$

Solution: Note that $$\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t}{t^2}dt=\ln 2=\lim_{x\to 0^+}\int_x^{2x}\frac{dt}{t}\\\iff\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t-t}{t^2}dt=0.$$

Thus it is sufficient to prove that $$\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t-t}{t^2}dt=0.$$

Now, expanding $\sin t$ using Maclaurin's formula having remainder in Lagrange's form, we have $$\sin t=t-\frac{\cos\xi}{3!}t^3, \text{ where } \xi=\theta t\text{ and } 0<\theta<1.$$

Now $$\cos \xi\le 1\implies \frac{\cos \xi}{3!}t^3\le \frac{t^3}{3!}(\because t>0)\\\implies t-\frac{\cos \xi}{3!}t^3\ge t-\frac{t^3}{3!}\\\implies \sin t\ge t-\frac{t^3}{3!}.$$

Again since $t>0$, we have $\sin t<t$.

Thus $\forall t\in[x,2x]$ and $\forall x>0,$ we have $$t-\frac{t^3}{3!}\le \sin t\le t\\\implies -\frac{t}{3!}\le \frac{\sin t-t}{t^2}\le 0\\\implies \int_x^{2x} -\frac{t}{3!}dt\le \int_x^{2x}\frac{\sin t-t}{t^2}dt\le 0\\\implies -\frac{x^2}{4}\le \int_x^{2x}\frac{\sin t-t}{t^2}dt\le 0.$$

Now since $\lim_{x\to 0^+} -\frac{x^2}{4}=0$ and $\lim_{x\to 0^+} 0=0$, thus by Sandwich Theorem we can conclude that $$\lim_{x\to 0^+} \int_x^{2x}\frac{\sin t-t}{t^2}dt=0.$$ Hence, we are done.

Is this solution correct and rigorous enough? Are there any alternative solutions?

Answer 1

There is an alternative solution. Integrating by parts with $u=\sin(t)$ and $v=-\frac1t$ we find that

$$\begin{align}\lim_{x\to 0^+}\int_x^{2x}\frac{\sin(t)}{t^2}\,dt&=\lim_{x\to 0^+}\left.\left(-\frac{\sin(t)}{t}\right)\right|_x^{2x}+\lim_{x\to 0^+}\int_x^{2x}\frac{\cos(t)}{t}\,dt\\\\ &=\lim_{x\to 0^+}\int_x^{2x}\frac{\cos(t)}{t}\,dt=\log(2)+\int_x^{2x}\frac{\cos(t)-1}{t}\,dt \end{align}$$

Then, note that for $t\in [x,2x]$, $|1-\cos(t)|=2\sin^2(t/2)\le t^2/2$. Therefore, $\lim_{x\to 0^+}\int_x^{2x}\frac{\cos(t)-1}{t}\,dt=0$ and we conclude

$$\lim_{x\to 0^+}\int_x^{2x}\frac{\sin(t)}{t^2}\,dt=\log(2)$$

as expected!

Answer 2

You can use the fact that $y = \dfrac{\sin x}{x}$ is positive and decreasing on $(0,\pi)$ so that if $0 < x < \pi/2$ you have $$ \frac{\sin 2x}{2x} \ln 2 = \frac{\sin 2x}{2x} \int_x^{2x} \frac 1t \, dt \le \int_x^{2x} \frac{\sin t}{t^2} \, dt \le \frac{\sin x}{x} \int_x^{2x} \frac 1t \, dt = \frac{\sin x}{x} \ln 2.$$ Now apply the squeeze theorem.

Answer 3

MVT for Integrals:

$\displaystyle{\int_{x}^{2x}}\dfrac{\sin t}{t}\frac{1}{t}dt =\dfrac{\sin s}{s}\int_{x}^{2x}\frac{1}{t}dt=$

$\dfrac{\sin s}{s}\left (\log 2x-\log x \right)$, where $s \in [x,2x]$.

Take the limit $x \rightarrow 0^+$.

Note: $\lim_{x \rightarrow 0^+} s=0,$ and

$\lim_{x \rightarrow 0^+}\dfrac{\sin s}{s}=1$.

Answer 4

I find it is easier to prove a generalization. Claim: Suppose $f$ is continuous on some $(0,a)$ and $\lim_{x\to 0^+} f(x) =1.$ Then

$$\lim_{x\to 0^+} \int_x^{2x} \frac{f(t)}{t}\, dt = \ln 2.$$

In our specific problem we have $f(t)= \dfrac{\sin t}{t}.$

Proof of claim:

$$\left | \int_x^{2x} \frac{f(t)}{t}\, dt - \ln 2\right | = \left |\int_x^{2x}\left (f(t)-1\right)\frac{dt}{t}\right |$$ $$\le \int_x^{2x}|f(t)-1|\frac{dt}{t} \le \max_{[x,2x]}|f-1|\cdot \int_x^{2x} \frac{dt}{t}$$ $$ = \max_{[x,2x]}|f-1| \cdot \ln 2.$$

Because $\lim_{x\to 0^+} f(x) =1,$ that maximum $\to 0$ as $x\to 0^+,$ and the claim is proved.

Answer 5

Let $ x\in\left(0,1\right] : $

\begin{aligned}\int_{x}^{2x}{\frac{\sin{t}}{t^{2}}\,\mathrm{d}t}&=\ln{2}+\int_{x}^{2x}{\frac{t-\sin{t}}{t^{2}}\,\mathrm{d}t}\\ &=\ln{2}+\int_{0}^{2x}{\frac{t-\sin{t}}{t^{2}}\,\mathrm{d}t}-\int_{0}^{x}{\frac{t-\sin{t}}{t^{2}}\,\mathrm{d}t}\\ &=\ln{2}+\frac{1}{2}\int_{0}^{x}{\frac{2u-\sin{\left(2u\right)}}{u^{2}}\,\mathrm{d}u}-\int_{0}^{x}{\frac{t-\sin{t}}{t^{2}}\,\mathrm{d}t}\\ \int_{x}^{2x}{\frac{\sin{t}}{t^{2}}\,\mathrm{d}t}&=\ln{2}+\int_{0}^{x}{\frac{\sin{t}\left(1-\cos{t}\right)}{t^{2}}\,\mathrm{d}t}\end{aligned}

Since $ x\mapsto\frac{x\left(1-\cos{x}\right)}{x^{2}} $ is continuous on $ \left(0,1\right] $, and can be extended to a continuous function $ \left[0,1\right] $, it can be upper-bounded on $ \left(0,1\right] $ by some constant $ M>0 :$ $$ \left(\exists M>0\right)\left(\forall x\in\left(0,1\right]\right),\ \frac{\sin{x}\left(1-\cos{x}\right)}{x^{2}}\leq M $$

Thus $ \int_{0}^{x}{\frac{\sin{t}\left(1-\cos{t}\right)}{t^{2}}\,\mathrm{d}t}\leq Mx\underset{x\to 0}{\longrightarrow}0 $, and hence : $$ \lim_{x\to 0}{\int_{x}^{2x}{\frac{\sin{t}}{t^{2}}\,\mathrm{d}t}}=\ln{2} $$

Answer 6

Put $t=ux$ to reduce the expression under limit to $$\int_{1}^{2}\frac{\sin ux} {u^2x}\,du$$ The integrand $$f(x, u) =\frac{\sin (ux)} {u^2x},f(0,u)=\frac{1}{u}$$ is continuous in $[0,h]\times[1,2]$ and hence we can switch limit with integral to get the desired limit as $$\int_{1}^{2}\frac{du}{u}=\log 2$$