TITLE: Evaluate $\lim_{R\to\infty}\left(\int_0^R\left|\frac{\sin x}{x}\right|dx-\frac{2}{\pi}\log R\right)$
QUESTION [6 upvotes]: Is there a closed form of 
$$\lim_{R\to\infty}\left(\int_0^R\left|\frac{\sin x}{x}\right|dx-\frac{2}{\pi}\log R\right)$$
I am pretty interested whether we can find out a closed form of this limit. 
We can show that for $R=n\pi,n\in\mathbb{N}$, we have
$$\begin{aligned}
\int_0^R\left|\frac{\sin x}{x}\right|dx&=\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{x}dx\\
&=\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin x}{(k+1)\pi-x}dx\\
&\leq \int_0^\pi\frac{\sin x}{\pi-x}dx+\sum_{k=1}^{n-1}\int_{0}^{\pi}\frac{\sin x}{k\pi}dx\\
&=\int_0^\pi\frac{|\sin(\pi-x)|}{x}dx+\sum_{k=1}^{n-1}\frac{2}{k\pi}dx\\
&=\int_0^\pi\frac{\sin x}{x}dx+\frac{2}{\pi}\sum_{k=1}^{n-1}\frac{1}{k}
\end{aligned}$$
On the other hand we have
$$\begin{aligned}
\int_0^R\left|\frac{\sin x}{x}\right|dx&\geq \sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin x}{(k+1)\pi}dx\\
&=\sum_{k=1}^n\frac{1}{k\pi}\int_0^\pi\sin xdx\\
&=\frac{2}{\pi}\sum_{k=1}^n\frac{1}{k}
\end{aligned}$$
Then I tried to apply the squeeze rule, but this does not lead to anything appetizing. Anybody know any tricks for this problem?

REPLY [2 votes]: Getting a closed form seems probably hard, if not impossible. But at least we can show the limit exists. This follows from the following inequality:
$$\left|\int_{k\pi}^{(k+1)\pi}\frac{|\sin(t)|}{t}-\frac2\pi\int_{k\pi}^{(k+1)\pi}\frac1t\right|\le\frac{c}{k^2}.\quad(1)$$
Which you prove by comparing both integrals to $\frac2{k\pi}$. For the first, $$\int_{k\pi}^{(k+1)\pi}\frac{|\sin(t)|}{t}-\frac2{k\pi}
=\int_{k\pi}^{(k+1)\pi}|\sin(t)|\left(\frac1t-\frac1{k\pi}\right).$$Now if $k\pi\le t\le(k+1)\pi$ then $$\left|\frac1t-\frac1{k\pi}\right|=\frac{t-k\pi}{k\pi t}\le\frac{1}{k^2\pi}.$$Inserting this above shows that $$\left|\int_{k\pi}^{(k+1)\pi}\frac{|\sin(t)|}{t}-\frac2{k\pi}\right|\le\frac2{k^2\pi}.\quad(2)$$
Similarly $$\left|\frac2\pi\int_{k\pi}^{(k+1)\pi}\frac1t-\frac2{k\pi}\right|\le\frac c{k^2},\quad(3)$$and then (1) follows from (2) and (3).