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Playing around with floor functions, I thought about using the maximum and minumum of functions to “merge” 2 asymptotic graphs which intersected each other infinitely. I then thought of the classical trigonometric Dirichlet Integral. Here is a graph of our goal integral. Please note the sign function:

$$\int_\Bbb R f(x)dx=\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+ \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx = \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)+ \max\left(-\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx =\int_0^\infty \frac{\cos(x) \text{sgn}\left(\frac{\sin(x)}x - \frac{\cos(x)}x\right)}{2 x} +\frac{\sin(x) \text{sgn}\left(\frac {\cos(x)}x - \frac{\sin(x)}x\right)}{2 x} + \frac{\sin(x)}{2 x} + \frac{\cos(x)}{2 x}-\frac{\cos(x) \text{sgn}\left(\frac{\sin(x)}x + \frac{\cos(x)}x\right)}{2 x}+ \frac{\sin(x) \text{sgn}\left(-\frac {\cos(x)}x - \frac{\sin(x)}x\right)}{2 x} + \frac{\sin(x)}{2 x} - \frac{\cos(x)}{2 x} dx =\frac\pi 2+ \int_0^\infty \frac{\cos(x) \text{sgn}\left(\frac{\sin(x)}x - \frac{\cos(x)}x\right)}{2 x} +\frac{\sin(x) \text{sgn}\left(\frac {\cos(x)}x - \frac{\sin(x)}x\right)}{2 x} -\frac{\cos(x) \text{sgn}\left(\frac{\sin(x)}x + \frac{\cos(x)}x\right)}{2 x}+ \frac{\sin(x) \text{sgn}\left(-\frac {\cos(x)}x - \frac{\sin(x)}x\right)}{2 x} dx=\frac\pi 2+\frac12 \int_0^\infty\frac{|\cos(x)+\sin(x)|}{x}-\frac{|\cos(x)-\sin(x)|}{x}dx=\boxed{ \frac\pi 2+\frac1{\sqrt 2} \int_0^\infty\frac{\left|\sin\left(x+\frac\pi 4\right)\right|}{x}-\frac{\left|\sin\left(x-\frac\pi 4\right)\right|}{x}dx }=\frac{\pi}{2}+ \lim_{x\to \infty}\left(\frac12\text{Ci}(x) \left[\text {sgn}\,\sin\left(x + \frac \pi4\right) -\text{sgn}\,\cos \left(x + \frac{\pi}4\right)\right] + \frac12 \text{Si}(x) \left[\text {sgn}\,\sin\left(x + \frac{\pi}4\right) + \text{sgn}\,\cos \left(x + \frac{\pi}4\right)\right]\right)\mathop=^{???}\pi$$

Our integral uses trigonometric integrals. I need to verify this solution. Here is a graph of that final integrand. Here is a graph of the actual problem; this plot is also a visual representation of our constant:

enter image description here

enter image description here

We used the Dirichlet Integral in one step. Here is a verification of the maximum and minimum function terms. Can you please verify this solution or give another way to evaluate it? Desmos and Wolfram Alpha cannot find a numerical approximation for the problem. Please correct me and give feedback!

Similar results which should have no typos:

$$\int_\Bbb R \min \left(\frac{\sin(x)}{x},e^{-|x|}\right)dx=2$$

$$\int_\Bbb R \max\left(\frac{\sin(x)}{x},e^{-|x|}\right)dx=\pi$$

Final result using Silver Ratio $\delta_S$:

$$\int_\Bbb R f(x)dx=\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+ \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx=\frac\pi 2+\coth^{-1}\left(\sqrt 2\right)=\frac\pi 2+\ln\left(\sqrt 2+1\right)=\frac\pi 2 +\ln(\delta_S)= 2.452169913814439644463931016619543751126745027949188…$$

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  • $\begingroup$ Sorry @amWhy, but the edited parentheses you made smaller do not fit the size of the fractions. $\endgroup$ Sep 22, 2021 at 17:33

2 Answers 2

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I will concentrate on this part of your proof, the calculation of this integral (the other steps of your proof are correct):

$$\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx$$

We will make use of the Lobachevsky integral formula:

Let $f(x)$ a $\pi$-periodic function (continous or integrable over its period). with

$\displaystyle f(x+\pi) = f(x)$ and $f(\pi-x) = f(x)$, for $0\leq x <\infty$. Then

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} f(x) dx = \int_{0}^{\frac{\pi}{2}} f(x) dx. $

Now, for your integral: If you make the change of variable $\displaystyle x=\frac{w}{2}$

$\displaystyle I=\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx = \int_{0}^{\infty} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{w} dw $

Multiplying and dividing by $\sin w$:

$$I= \int_{0}^{\infty} \frac{\sin w}{w} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w} dw$$

The function

$$f(w) = \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}$$

is $\pi$-periodic

By the Lobachevsky integral formula:

$$\displaystyle I=\int_{0}^{\infty} \frac{\sin w}{w} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w} dw = \int_{0}^{\frac{\pi}{2}}\frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}dw $$

Note that for $w\in\left(0,\frac{\pi}{2}\right)$

$$\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big|= \sin\left(\frac{w}{2}+\frac{\pi}{4}\right)$$

$$\Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|= -\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)$$

Hence

$$\displaystyle\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big| = \sqrt{2}\sin\left(\frac{w}{2}\right)\displaystyle $$ Therefore

$$ I = \int_{0}^{\frac{\pi}{2}}\frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}dw = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{w}{2}\right)}{\sin w}dw$$

Using $\displaystyle \frac{\sin\left(\frac{w}{2}\right)}{\sin w} = \frac{1}{2}\sec\left(\frac{w}{2}\right)$

we have

$$I = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{w}{2}\right)}{\sin w}dw = \frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}} \sec\left(\frac{w}{2} \right)dw = \sqrt{2} \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)$$

We can conclude

$$\boxed{\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx = \sqrt{2} \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)}$$

Therefore

$$\boxed{\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+ \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx = \frac{\pi}{2} +\operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)} $$

Some numerical approximations from Wolfram:

Integral with max part

Integral with min part

The sum is 2.44206

while

$$\frac{\pi}{2} + \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)\approx 2.45216..$$

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    $\begingroup$ Does this mean that $$\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+\int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx=\frac\pi 2+\coth^{-1}\ \sqrt 2?$$ $\endgroup$ Sep 22, 2021 at 15:52
  • $\begingroup$ Need to review the other steps. But at first sight seems that this is the answer. I thought your main question was about the integral I solved. $\endgroup$
    – Bertrand87
    Sep 22, 2021 at 16:00
  • $\begingroup$ Your answer was just the integral I got using this formula. Please see the graph of the constant. Thanks again.- $\endgroup$ Sep 22, 2021 at 16:01
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    $\begingroup$ @zinoviev Nice solution. I get $\sqrt{2}\ln(1 + \sqrt{2})$ which is the same as yours. $\endgroup$
    – River Li
    Sep 22, 2021 at 16:27
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    $\begingroup$ @TymaGaidash Great problem! $\endgroup$
    – Bertrand87
    Sep 22, 2021 at 18:00
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Problem: Prove that $$\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} \,\mathrm{d}x = \sqrt{2}\ln(1 + \sqrt{2}).$$ Remark: We have $\lim_{x\to 0}\frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} = \sqrt2$.

Sketch of an alternative proof:

Using the identity
$$\left|\sin x\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k x),$$ we have $$\frac{|\sin(x + \pi/4)| - |\sin(x - \pi/4)|}{x} = \sum_{k=1}^\infty \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\, \frac{\sin (2kx)}{x}.$$

Let \begin{align*} f(N) &= \int_0^N \frac{|\sin(x + \pi/4)| - |\sin(x - \pi/4)|}{x}\,\mathrm{d} x, \\ g(N) &= \int_0^N \sum_{k=1}^N \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\, \frac{\sin (2kx)}{x}\, \mathrm{d} x. \end{align*}

First, we have \begin{align*} g(N) &= \sum_{k=1}^N \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)} \int_0^N \frac{\sin (2kx)}{x}\, \mathrm{d} x\\ &= \sum_{k=1}^N \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\int_0^{2kN} \frac{\sin y}{y}\,\mathrm{d} y\\ &= \sum_{k=1}^N \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\cdot\frac{\pi}{2} - \sum_{k=1}^N \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\int_{2kN}^\infty \frac{\sin y}{y}\,\mathrm{d} y \end{align*} where we have used $\int_0^\infty \frac{\sin y}{y}\,\mathrm{d} y = \frac{\pi}{2}$. Then, one can prove that $$\lim_{N\to \infty} g(N) = \sum_{k=1}^\infty \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\cdot \frac{\pi}{2} = \sum_{m=1}^\infty \frac{4 (-1)^{m + 1}}{4(2m - 1)^2 - 1} = \sqrt{2}\ln(1 + \sqrt{2}).$$

Second, we have \begin{align*} f(N) - g(N) &= \int_0^N \sum_{k=N+1}^\infty \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\, \frac{\sin (2kx)}{x}\, \mathrm{d} x\\ &= \int_0^{1/\sqrt{N}} \sum_{k=N+1}^\infty \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\, \frac{\sin (2kx)}{x}\, \mathrm{d} x + \int_{1/\sqrt{N}}^N \sum_{k=N+1}^\infty \frac{8\sin\frac{k\pi}{2}}{\pi(4k^2 - 1)}\, \frac{\sin (2kx)}{x}\, \mathrm{d} x\\ &\to 0, \quad \mathrm{as}\quad N \to \infty. \end{align*}

Thus, we have $\lim_{N\to \infty} f(N) = \lim_{N\to \infty} g(N) = \sqrt{2}\ln(1 + \sqrt{2})$.

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  • $\begingroup$ [+1] This is a good answer, but the newer answerer has a simpler method. $\endgroup$ Sep 29, 2021 at 13:21
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    $\begingroup$ @TymaGaidash Thanks. By the way, $\left|\sin x\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k x)$ is a useful identity. $\endgroup$
    – River Li
    Sep 29, 2021 at 14:52
  • $\begingroup$ How do you derive it? $\endgroup$ Sep 29, 2021 at 14:53
  • $\begingroup$ @TymaGaidash It is a known result. I just used it, $\endgroup$
    – River Li
    Sep 29, 2021 at 16:36

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