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Prove $$\int_{0}^{\infty}\frac{|\sin x|\sin x}{x}dx=1.$$ I know how to calculate $\int_{0}^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}$, but the method cannot be applied here. So I am thinking $$\sum_{k=0}^n(-1)^k\int_{k\pi}^{(k+1)\pi}\frac{\sin^2 x}{x}dx$$ but I don't know how to proceed.

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  • $\begingroup$ You have $$\int_0^\pi\,\frac{\sin^2(x)}{x}\,\text{d}x=\frac12\,\big(\gamma+\ln(2\pi)-\text{Ci}(2\pi)\big)$$ and $$\int_{k\pi}^{(k+1)\pi}\,\frac{\sin^2(x)}{x}\,\text{d}x=\frac12\,\Big( \text{Ci}(2k\pi)-\text{Ci}\big(2(k+1)\pi\big)+\ln(k+1)-\ln(k)\Big)$$ for integers $k=1,2,3,\ldots$. Here, $\gamma$ is the Euler–Mascheroni constant, and $\text{Ci}$ is the cosine integral: $$\text{Ci}(z)=-\int_z^\infty\,\frac{\cos(t)}{t}\,\text{d}t\,.$$ $\endgroup$ – Batominovski Oct 12 '18 at 11:43
  • $\begingroup$ That is, $$\int_0^\infty\,\frac{\sin(x)\,\big|\sin(x)\big|}{x}\,\text{d}x=\frac{1}{2}\,\big(\gamma+\ln(2\pi)\big)+\sum_{k=1}^\infty\,(-1)^k\,\text{Ci}(2k\pi)+\frac12\,\sum_{k=1}^\infty\,(-1)^{k}\,\ln\left(1+\frac{1}{k}\right)\,.$$ I am not sure whether this is helpful. Mathematica seems to confirm that $$\frac{1}{2}\,\big(\gamma+\ln(2\pi)\big)+\sum_{k=1}^\infty\,(-1)^k\,\text{Ci}(2k\pi)+\frac12\,\sum_{k=1}^\infty\,(-1)^{k}\,\ln\left(1+\frac{1}{k}\right)=1\,.$$ $\endgroup$ – Batominovski Oct 12 '18 at 11:43
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By Lobachevsky integral formula: https://en.wikipedia.org/wiki/Lobachevsky_integral_formula $$\int_{0}^{\infty}\frac{\sin x}{x}|\sin x|\,\mathrm{d}x=\int_0^{\pi/2}|\sin x|\,\mathrm{d}x=1.$$

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  • $\begingroup$ +1. $\texttt{Mathematica}$ doesn't seem to know the Lobachevsky stuff. $\endgroup$ – Felix Marin Oct 12 '18 at 19:49
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We have $$ \left|\sin x\right|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1} \tag{1} $$ $$\forall n\in\mathbb{N}^+,\quad \int_{0}^{+\infty}\frac{\sin(x)\cos(2nx)}{x}\,dx=0\tag{2} $$ hence $$ \int_{0}^{+\infty}\frac{\left|\sin x\right|\sin x}{x}\,dx = \frac{2}{\pi}\int_{0}^{+\infty}\frac{\sin x}{x}\,dx = 1.\tag{3}$$

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  • $\begingroup$ (+1) Nicely done Jack. No need to appeal to anything other that Fourier Series representation of $| \sin(x)|$ and the value of $\int_0^\infty \frac1{\sin(x)}{x}\,dx$. $\endgroup$ – Mark Viola Oct 12 '18 at 20:17
  • $\begingroup$ Why is $\int_{0}^{+\infty}\frac{\sin(x)\cos(2nx)}{x}\,dx=0$ ? (some reference would do - thanks) $\endgroup$ – Andreas Oct 15 '18 at 12:34
  • $\begingroup$ @Andreas: because $\int_{0}^{+\infty}\frac{\sin(m x)}{x}\,dx = \frac{\pi}{2}$ for any $m\in\mathbb{N}^+$ and you may invoke the sine addition formulas. $\endgroup$ – Jack D'Aurizio Oct 15 '18 at 12:35
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Let $$I:=\int_0^\infty\,\frac{\big|\sin(x)\big|\,\sin(x)}{x}\,\text{d}x\,.$$ Therefore, $$2I=\int_{-\infty}^{+\infty}\,\frac{\big|\sin(x)\big|\,\sin(x)}{x}\,\text{d}x=\int_0^\pi\,\sin^2(x)\,\left(\sum_{k=-\infty}^{+\infty}\,\frac{(-1)^k}{x+k\pi}\right)\,\text{d}x\,.$$ It can be proven by residue calculus that $$\text{csc}(z)=\sum_{k=-\infty}^{+\infty}\,\frac{(-1)^k}{z+k\pi}\text{ for all }z\in\mathbb{C}\setminus\pi\mathbb{Z}\,.$$ Thus, $$2I=\int_0^\pi\,\sin^2(x)\,\text{csc}(x)\,\text{d}x=\int_0^\pi\,\sin(x)\,\text{d}x=2\,,$$ whence $I=1$.

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    $\begingroup$ Indeed, the proof of the Lobachevsky integral formula seems to be along similar lines math.stackexchange.com/questions/776903/… $\endgroup$ – Calvin Khor Oct 12 '18 at 12:00
  • $\begingroup$ @CalvinKhor Oh, nice! I didn't realize that this proof can be extended to prove the Lobachesky Integral Formula. $\endgroup$ – Batominovski Oct 12 '18 at 12:01
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    $\begingroup$ @batominovski (+1) One need not use residue calculus to develop the partial fraction representation of the cosecant function, but can develop the representation using real analysis only. See THE APPENDIX OF THIS ANSWER $\endgroup$ – Mark Viola Oct 12 '18 at 20:23

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