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?

  • $\begingroup$ That limit does not exist (well actually it equals $-\infty$, if you want to count that as existing). You wouldn't expect it to, because the average (loosely speaking) off $|\sin(t)|$ is not $1$. In fact $\int_{k\pi}^{(k+1)\pi}|\sin(t)|=2$, so what I suspect exists is the limit with $\frac2\pi\log(R)$ in place of $\log(R)$. $\endgroup$ – David C. Ullrich Dec 16 '15 at 22:20
  • $\begingroup$ @DavidC.Ullrich You're absolutely correct,there should be a factor $\frac{2}{\pi}$ before $\log R$. I've updated the post. $\endgroup$ – Frank Lu Dec 16 '15 at 22:23
  • $\begingroup$ @AndréNicolas I've edit the post, there should be a factor $\frac{2}{\pi}$ before $\log R$. $\endgroup$ – Frank Lu Dec 16 '15 at 22:23

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).


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