Evaluate $\lim_{R\to\infty}\left(\int_0^R\left|\frac{\sin x}{x}\right|dx-\frac{2}{\pi}\log R\right)$

Question

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?

Answer 1

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).

Answer 2

We have \begin{align*} \int_0^{\pi N} {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} & = \frac{2}{\pi }\sum\limits_{k = 1}^N {\frac{1}{k}} + \sum\limits_{k = 1}^N {\int_0^\pi {\left[ {\frac{{\sin x}}{{\pi k - x}} - \frac{{\sin x}}{{\pi k}}} \right]{\rm d}x} } \\ & = \frac{2}{\pi }\sum\limits_{k = 1}^N {\frac{1}{k}} + \sum\limits_{k = 1}^N {\int_0^\pi {\frac{{x\sin x}}{{\pi k(\pi k - x)}}{\rm d}x} } . \end{align*} It is not difficult to show that the second sum converges as $N\to+\infty$ and the error committed by stopping at the $N$th term is $\mathcal{O}(N^{-1})$. Consequently, by using the standard approximation for the harmonic numbers, \begin{align*} \int_0^{\pi N} {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} & = \frac{2}{\pi }\log N + \frac{2}{\pi }\gamma + \sum\limits_{k = 1}^\infty {\int_0^\pi {\frac{{x\sin x}}{{\pi k(\pi k - x)}}{\rm d}x} } + \mathcal{O}\!\left( {\frac{1}{N}} \right) \\ & = \frac{2}{\pi }\log N + \frac{2}{\pi }\gamma + \sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{x\sin (\pi x)}}{{k(k - x)}}{\rm d}x} } + \mathcal{O}\!\left( {\frac{1}{N}} \right) \end{align*} as $N\to +\infty$ ($\gamma$ being the Euler–Mascheroni constant). It is easy to see that $$ \int_0^R {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} - \int_0^{\pi \left\lfloor {R/\pi } \right\rfloor } {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} = \mathcal{O}\!\left( {\frac{1}{R}} \right), $$ whence $$\boxed{ \int_0^R {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} = \frac{2}{\pi }\log R + C + \mathcal{O}\!\left( {\frac{1}{R}} \right)} $$ as $R\to +\infty$, with $$\boxed{ C = \frac{2}{\pi }(\gamma - \log \pi ) + \sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{x\sin (\pi x)}}{{k(k - x)}}{\rm d}x} } .} $$ We may obtain a some alternative expressions for $C$ as follows. Note that \begin{align*} &\sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{x\sin (\pi x)}}{{k(k - x)}}{\rm d}x} } = \int_0^1 {\left( {\sum\limits_{k = 1}^\infty {\frac{x}{{k(k - x)}}} } \right)\sin (\pi x){\rm d}x}\\ & = - \int_0^1 {(\gamma + \psi (1 - x))\sin (\pi x){\rm d}x} = - \frac{{2\gamma }}{\pi } - \int_0^1 {\psi (1 - x)\sin (\pi x){\rm d}x} \\ & = - \frac{{2\gamma }}{\pi } - \int_0^1 {\psi (t)\sin (\pi t){\rm d}t} = - \frac{{2\gamma }}{\pi } + \pi \int_0^1 {\log \Gamma (t)\cos (\pi t){\rm d}t} . \end{align*} Here $\psi$ and $\Gamma$ are the digamma function and the gamma function, respectively. Consequently, $$\boxed{ C = - \frac{{2\log \pi }}{\pi } -\int_0^1 {\psi (t)\sin (\pi t){\rm d}t}= - \frac{{2\log \pi }}{\pi } + \pi \int_0^1 {\log \Gamma (t)\cos (\pi t){\rm d}t} .} $$ Using the Fourier series of $\log\Gamma(t)$, we find \begin{align*} \int_0^1 {\log \Gamma (t)\cos (\pi t){\rm d}t} & = \frac{{2(\gamma + \log (2\pi ))}}{{\pi ^2 }} + \frac{4}{{\pi ^2 }}\sum\limits_{n = 1}^\infty {\frac{{\log n}}{{4n^2 - 1}}} \\ & = \frac{{2(\gamma + \log (2\pi ))}}{{\pi ^2 }} + \frac{4}{{\pi ^2 }}\sum\limits_{n = 1}^\infty {\frac{{\log n}}{{(2n - 1)(2n + 1)}}} \\ & = \frac{{2(\gamma + \log (2\pi ))}}{{\pi ^2 }} - \frac{2}{{\pi ^2 }}\sum\limits_{n = 2}^\infty {\frac{{\log (1 - 1/n)}}{{2n - 1}}} . \end{align*} Hence, $$\boxed{ C =\frac{2}{\pi }\left[ {\gamma + \log 2 +2 \sum\limits_{n = 1}^\infty {\frac{{\log n}}{{4n^2 - 1}}}} \right]= \frac{2}{\pi }\left[ {\gamma + \log 2 - \sum\limits_{n = 2}^\infty {\frac{{\log (1 - 1/n)}}{{2n - 1}}} } \right].} $$ Numerically, $C= 1.1129238102\ldots$. Note that $$ \int_0^1 {\psi (t)\sin (\pi t){\rm d}t} = \int_0^1 {\psi (t + 1)\sin (\pi t){\rm d}t} - \int_0^\pi {\frac{{\sin t}}{t}{\rm d}t} , $$ and the latter is easier to compute numerically.

Answer 3

An alternative way to get $L=\frac2\pi(\gamma+\log2)+\frac4\pi\sum_{n=1}^\infty\frac{\log n}{4n^2-1}$ for the limit, written as $$L=\int_0^1|\sin x|\,\frac{dx}{x}-\int_1^\infty\left(\frac2\pi-|\sin x|\right)\frac{dx}{x},$$ is to use the Fourier series $$|\sin x|=\frac2\pi-\frac4\pi\sum_{n=1}^\infty\frac{\cos 2nx}{4n^2-1}=\frac4\pi\sum_{n=1}^\infty\frac{1-\cos 2nx}{4n^2-1},$$ which gives $L=\frac4\pi\sum_{n=1}^\infty\frac{f(2n)}{4n^2-1}$ with known $$f(a)=\int_0^1\frac{1-\cos ax}{x}\,dx-\int_1^\infty\frac{\cos ax}{x}\,dx=\gamma+\log a.$$