Convergence of a special integral Determine the value of the integral below:
$\lim_{x\rightarrow 0^+} \int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ $
My attempt to solution:
I could not solve the problem. And I do not know how to continue.
$1)$ I started analysing just  integral, and after that I applied the integration by parts:
$\int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ = -\frac{\sin{(2x)}}{2x} +\frac{\sin{(x)}}{x} - \int_{x}^{2x} \frac{-\cos{(t)}}{t} \,dt\ $
$\int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ = -\frac{\sin{(2x)}}{2x} +\frac{\sin{(x)}}{x} + \int_{0}^{2x} \frac{\cos{(t)}}{t} \,dt\ - \int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ $
$2)$ Now, applying the limit:
$\lim_{x\rightarrow +0} \int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ = \lim_{x\rightarrow +0} (-\frac{\sin{(2x)}}{2x} +\frac{\sin{(x)}}{x} + \int_{0}^{2x} \frac{\cos{(t)}}{t} \,dt\ - \int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ )$
$3)$ If we prove that $\exists\lim_{x\rightarrow +0} \frac{\sin{(x)}}{x}$, $\exists\lim_{x\rightarrow +0}\int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ $, we could be able to apply the sum of limits:
$\lim_{x\rightarrow +0} \int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ = \lim_{x\rightarrow +0} -\frac{\sin{(2x)}}{2x} +\lim_{x\rightarrow +0} \frac{\sin{(x)}}{x} + \lim_{x\rightarrow +0}\int_{0}^{2x} \frac{\cos{(t)}}{t} \,dt\ - \lim_{x\rightarrow +0} \int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ $
$4)$ I know that $\lim_{x\rightarrow +0}\frac{\sin{(x)}}{x} = 1$, but I do not know if $ \exists\lim_{x\rightarrow +0} \int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ $, and how to find the value of it.
I tried for long time, but I could not find.
 A: In the limit as $x\to0^+$, the following two effects compete each other:

*

*The interval of integration, $[x, 2x]$ shrinks.


*The average magnitude of the integrand $\frac{\sin t}{t^2}$ for $x \leq t \leq 2x$ increases.
One way to cancel out these two effects is to substitute $t=xu$:
$$ \int_{x}^{2x} \frac{\sin t}{t^2} \, \mathrm{d}t
= \int_{1}^{2} \frac{\sin (xu)}{x u^2} \, \mathrm{d}u. $$
In light of this and $\lim_{r \to 0} \frac{\sin r}{r} = 1$ together, it is natural to expect that the limit is $\int_{1}^{2} \frac{1}{u} \, \mathrm{d}u = \log 2$. To justify this, for any $\varepsilon > 0$, choose $\delta > 0$ such that
$$ \left| r - 0 \right| < 2\delta \quad \Rightarrow \quad \left| \frac{\sin r}{r} - 1 \right| < \frac{\varepsilon}{\log 2}. $$
Now let $0 < x < \delta$. Then for any $u \in [1, 2]$, we have $0 < xu < 2\delta$, and so,
$$ \left| \frac{\sin(xu)}{xu} - 1 \right| < \frac{\varepsilon}{\log 2}. $$
Plugging this to the integral representation above,
\begin{align*}
\left| \int_{1}^{2} \frac{\sin (xu)}{x u^2} \, \mathrm{d}u - \log 2 \right|
&= \left| \int_{1}^{2} \frac{1}{u} \left( \frac{\sin (xu)}{xu} - 1 \right) \, \mathrm{d}u \right|
< \int_{1}^{2} \frac{1}{u} \cdot \frac{\varepsilon}{\log 2} \, \mathrm{d}u
= \varepsilon.
\end{align*}
Summarizing, we have established the $\varepsilon$-$\delta$ property for the limit $\lim_{x\to0^+} \int_{x}^{2x} \frac{\sin t}{t^2} \, \mathrm{d}t = \log 2$ and therefore we are done.
A: As the interval $[x, 2x]$ gets smaller and smaller the integrand will be $1/t + O(t)$, so the integral is $\log 2x - \log x + O(x^2)$, giving you $\log 2$ as the result as $x\to 0^+$.
A: Since $\cos x \to 1$ as $x \to 0+$, for any $\epsilon > 0$ there exists $\delta > 0$ such that $1- \epsilon \leqslant \cos t \leqslant 1$ for $0 < x < t< 2x < \delta$, and
$$(1-\epsilon) \log 2 = (1-\epsilon)\int_x ^{2x} \frac{dt}{t} \leqslant \int_x^{2x} \frac{\cos t}{t} \, dt \leqslant\int_x ^{2x} \frac{dt}{t}= \log 2$$
Thus, for all $x < \delta/2$,
$$\left|\int_x^{2x} \frac{\cos t}{t} \, dt - \log 2 \right| < \epsilon \log 2,$$
which implies
$$\lim_{x \to 0+}\int_x^{2x} \frac{\cos t}{t}\, dt = \log 2,$$
and from your attempt with integration by parts,
$$\lim_{x \to 0+}\int_x^{2x} \frac{\sin t}{t^2}\, dt = \underbrace{-\lim_{x \to 0+}\frac{\sin 2x}{2x} + \lim_{x \to 0+} \frac{\sin x }{x}}_{=-1+1=0} + \lim_{x \to 0+}\int_x^{2x} \frac{\cos t}{t}\, dt = \log 2$$
