How to solve the following types of improper integral problem? 1.Find the limit.
\begin{equation*}
\lim _{R\rightarrow \infty }( \ \int ^{2R}_{2\pi } \ \ \frac{\sin x}{x} dx\ -\ \int ^{R}_{\pi } \ \ \left(\frac{\sin x}{x}\right)^{2} dx
\end{equation*}
2.if  \begin{equation*}{R}≧{2\pi } \end{equation*} ,
Prove the following.
\begin{equation*}
\int ^{R}_{\pi } \ \ \left(\frac{\sin x}{x}\right)^{2} dx
=\frac{\sin 2R}{4R}+\frac{1}{2}\int ^{2R}_{2 \pi } \ \ \left(\frac{x-\sin x}{x^2}\right) dx
\end{equation*}
This is my upcoming entrance examination question.It's too difficult for me ......,
 A: Hints
The equality you want to prove comes from integrating by parts. The term outside integrals comes from a double arc identity for the sine ($\sin(2x) = 2\sin(x)\cos(x)$). The term that was integrated was 1.
Then, if you have the result, try replacing it on the limit you want to find.
A: Noting
$$ \int_{2\pi}^{2R}\frac{\sin x}{x}dx=\int_{\pi}^{R}\frac{\sin (2x)}{x}dx=-\frac{\cos(2x)}{2x}\bigg|_\pi^R-\frac12\int_{\pi}^{R}\frac{\cos (2x)}{x^2}dx $$
one has
\begin{eqnarray}
&&\lim _{R\rightarrow \infty }\left(\int ^{2R}_{2\pi } \frac{\sin x}{x} dx-\int ^{R}_{\pi }\left(\frac{\sin x}{x}\right)^{2} dx\right)\\
&=&\lim _{R\rightarrow \infty }\left( \frac1{2\pi}-\frac12\int_{\pi}^{R}\frac{\cos (2x)}{x^2}dx - \int ^{R}_{\pi } \left(\frac{\sin x}{x}\right)^{2} dx\right)\\
&=&\frac{1}{2\pi}-\lim _{R\rightarrow \infty }\frac12\int_{\pi}^{R}\frac{\cos (2x)+2\sin^2x}{x^2}dx\\
&=&\frac{1}{2\pi}-\lim _{R\rightarrow \infty }\frac12\int_{\pi}^{R}\frac{1}{x^2}dx\\
&=&\frac{1}{2\pi}-\frac{1}{2\pi}\\
&=&0.
\end{eqnarray}
