# Existence and the value of $\lim_{x \to 0} \int^{3x}_x \frac{\sin t}{t^2}dt$

I need the show the existence of the following limit and then calculate the limit

$$\lim_{x \to 0} \int^{3x}_x \frac{\sin t}{t^2}dt$$

Since the antiderivative of $$\frac{\sin t}{t^2}$$ was not nice, I tried to use the approximation $$\sin x \approx x$$ for $$x$$ close to $$0$$. Then, I can integrate and find the limit as $$\ln3$$.

Is this a valid solution and can I solve this without using such approximation?

Hint: You can apply an integration for parts taking $$u=\sin t$$ and $$v'=1/t^2$$ and you will found

$$=\left[-\frac{\sin t}{t}-\int -\frac{\cos t}{t}dt\right]^{3x}_x$$

Remember that exist the cosine integral named $$\operatorname{Ci}(x)$$ https://en.wikipedia.org/wiki/Trigonometric_integral:

$$\operatorname{Ci}(x) = -\int_x^\infty \frac{\cos t}{t}dt = \gamma + \ln x - \int_0^x \frac{1 - \cos t}{t}dt$$

By using the substitution $$u = \frac{t}{x}$$, you get $$\int^{3x}_x\frac{\sin(t)}{t^2}dt = \int^3_1\frac{\sin(ux)}{u^2x^2}\cdot x \cdot du = \int^3_1\frac{\sin(ux)}{u^2x}du$$ Now, since $$\left|\frac{\sin(ux)}{u^2x}\right| \leq \left|\frac{1}{u^2x}\right|$$ which is integrable and bounded on $$u \in [1, 3]$$, you can use dominated convergence and swap the limit and the integral. For the inner limit, note that by using l'Hospital's rule it follows that $$\lim_{x\rightarrow 0} \frac{\sin(ux)}{u^2x} = \lim_{x\rightarrow 0} \frac{u\cdot \cos(ux)}{u^2} = \frac{1}{u}.$$ Plugging this back into the integral yields: $$\lim_{x\rightarrow 0} \int^{3x}_x \frac{\sin(t)}{t^2}dt = \int^3_1 \frac{1}{u}du = \ln(3) - \ln(1) = \ln(3)$$

MVT for integrals

$$((\sin s)/s)\displaystyle{\int_{x}^{3x}}(1/t)dt=$$

$$((\sin s)/s)[\log 3+\log x -\log x] ;$$

where $$s \in [x,3x].$$

Note $$\lim x \rightarrow 0$$ implies $$\lim s \rightarrow 0$$.

Take the limit.

• I have upvoted before. Please, can you explain the significance of MVT? I'm with a low level in English language. Commented Jan 7, 2021 at 22:06
• @Sebastiano It stands for the Mean Value Theorem for integrals. Here is a link: en.wikipedia.org/wiki/… Commented Jan 7, 2021 at 22:45
• Sebastiano. The link given by C Squared should help. A quite well known theorem. If any more questions, just write. You will find this theorem in any introductory Analysis book. Commented Jan 8, 2021 at 7:56
• @PeterSzilas Hi and thank you very much for your explanation on MVT. I kwown very well the theorem but I have not understood the acronymus MVT. +1 for the comments :-) Commented Jan 8, 2021 at 11:33
• Sebastiano. In quite a few problesms with limits, where an integral is involved this can be of use.Greetings, peter Commented Jan 8, 2021 at 12:01

Since $$\sin(t)=t+O(t^3)$$,

$$\int^{3x}_x \frac{\sin t}{t^2}dt =\int^{3x}_x \frac{t+O(t^3)}{t^2}dt\\ =\int^{3x}_x (\frac1{t}+O(t))dt\\ =\ln(3)+O(x^2)\\ \to \ln(3)\\$$

You can use $$t-t^3/6 < \sin(t) < t$$ to get explicit bounds for the integral.

• Approved....:-) Commented Jan 7, 2021 at 21:04

Notice that, for any $$x\in\mathbb{R}$$; : \begin{aligned} \int_{x}^{3x}{\frac{\sin{y}}{y^{2}}\,\mathrm{d}y}&=\ln{3}-\int_{x}^{3x}{\frac{y-\sin{y}}{y^{2}}\,\mathrm{d}y}\\ &=\ln{3}-\int_{0}^{3x}{\frac{y-\sin{y}}{y^{2}}\,\mathrm{d}y}+\int_{0}^{x}{\frac{y-\sin{y}}{y^{2}}\,\mathrm{d}y}\\ &=\ln{3}-\int_{0}^{x}{\frac{3y-\sin{\left(3y\right)}}{27y^{2}}\,\mathrm{d}y}+\int_{0}^{x}{\frac{y-\sin{y}}{y^{2}}\,\mathrm{d}y}\\&=\ln{3}+\int_{0}^{x}{f\left(x\right)\mathrm{d}x} \end{aligned}

In the second line we were able to split the integral apart because $$x\mapsto\frac{x-\sin{x}}{x^{2}}$$ is piecewise continuous on any segment $$\left[a,b\right]\subset\left[0,+\infty\right)$$.

In the last line $$f$$ is non other than the function $$x\mapsto\frac{x-\sin{x}}{x^{2}}-\frac{3x-\sin{\left(3x\right)}}{27x^{3}}$$, which is also piecewise continuous on $$\left[0,a\right]$$, for any $$a>0$$.

Being piecewise continuous makes $$f$$ bounded on any segment $$\left[a,b\right] \subset\left[0,+\infty\right)$$, thus : $$\int_{0}^{x}{f\left(y\right)\mathrm{d}y}\underset{x\to 0}{\longrightarrow}0$$

Hence : $$\int_{x}^{3x}{\frac{\sin{y}}{y^{2}}\,\mathrm{d}y}\underset{x\to 0}{\longrightarrow}\ln{3}$$

• I wanted to give the same argument and then saw your answer. +1 Commented Jan 8, 2021 at 2:54