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?
 A: 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:
$$\operatorname{Ci}(x) = -\int_x^\infty \frac{\cos t}{t}dt = \gamma + \ln x - \int_0^x \frac{1 - \cos t}{t}dt$$
A: 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:
$$\lim_{x\rightarrow 0} \int^{3x}_x \frac{\sin(t)}{t^2}dt = \int^3_1 \frac{1}{u}du = \ln(3) - \ln(1) = \ln(3)$$
A: 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.
A: 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.
A: 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 : $$ \int_{x}^{3x}{\frac{\sin{y}}{y^{2}}\,\mathrm{d}y}\underset{x\to 0}{\longrightarrow}\ln{3} $$
