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I'm trying to calculate the following integral: $$ \int_0^\infty\frac{(-1)^{\lfloor x\rfloor}\sin(2\beta x)}{x}\mathrm dx, $$ where $\lfloor x\rfloor$ is the floor function. My first step was: $$\sum_{n=1}^\infty (-1)^n\int_n^{n+1}\frac{\sin(2\beta x)}{x}\mathrm dx $$ Since $\displaystyle \int_n^{n+1}\frac{\sin(2\beta x)}{x}\mathrm dx =\text{Si}(2\beta(n+1))-\text{Si}(2\beta n)$, where $\text{Si}(z)$ is the Sine Integral. Then we have the sum: $$\sum_{n=1}^\infty (-1)^n\left[\text{Si}(2\beta(n+1))-\text{Si}(2\beta n) \right] $$ Here is my problem, I have no idea how to calculate this sum. Perhaps the sum does not fit in terms of Sine Integral, I tried to take the sum into the integral before obtaining Sine Integral, but I was not successful. I will be grateful for help.

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    $\begingroup$ It should be: $$\int_0^{\infty } \frac{(-1)^{\lfloor x\rfloor } \sin (2 \beta x)}{x} \, dx=\text{Si}(2 \beta )+\sum _{n=1}^{\infty } (-1)^n (\text{Si}(2 \beta (n+1))-\text{Si}(2 \beta n))$$ $\endgroup$ – Mariusz Iwaniuk Dec 24 '20 at 10:23
  • $\begingroup$ as $\beta\to\infty$ integral tends to $\pi/2$ $\endgroup$ – Raffaele Dec 24 '20 at 13:17
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$$I(\beta)=\sum_{n=1}^\infty (-1)^n\Big[\text{Si}(2\beta(n+1))-\text{Si}(2\beta n) \Big]$$ $$I'(\beta)=\sum_{n=1}^\infty (-1)^n\Big[2 (n+1) \text{sinc}(2 \beta (n+1))-2 n \text{sinc}(2 \beta n)\Big]$$ Given by a CAS $$I'(\beta)=\frac{\tan (\beta )-\sin (3 \beta ) \sec (\beta )}{2 \beta }$$ $$I(\beta)=\int\frac{\tan (\beta )-\sin (3 \beta ) \sec (\beta )}{2 \beta }\, d\beta$$ As far as I can see, there is no closed form for the antiderivative.

If $\beta < \frac \pi 2$, we could make a series expansion around $\beta=0$ $$\frac{\tan (\beta )-\sin (3 \beta ) \sec (\beta )}{2 \beta }=\sum_{n=0}^\infty (-1)^n 2^{2n+1}\frac{E_{2 n+1}(-1)+1}{(2n+1)!}\, \beta^{2n}$$ $$I(\beta)=\sum_{n=0}^\infty (-1)^n 2^{2n+1}\frac{E_{2 n+1}(-1)+1}{(2n+1)^2\,(2n)!}\, \beta^{2n+1}$$

Limiting the summation to $1000$ terms, some results $$\left( \begin{array}{ccc} \beta & \text{summation} & \text{integration} \\ 0.1 & -0.0994447100 & -0.0994447100 \\ 0.2 & -0.1955639426 & -0.1955639426 \\ 0.3 & -0.2850622742 & -0.2850622742 \\ 0.4 & -0.3646983033 & -0.3646983033 \\ 0.5 & -0.4312954393 & -0.4312954393 \\ 0.6 & -0.4817299506 & -0.4817299506 \\ 0.7 & -0.5128815984 & -0.5128815984 \\ 0.8 & -0.5215216327 & -0.5215216325 \\ 0.9 & -0.5040901387 & -0.5040901367 \\ 1.0 & -0.4562617463 & -0.4562617462 \\ 1.1 & -0.3720624328 & -0.3720624322 \\ 1.2 & -0.2418977415 & -0.2418977414 \\ 1.3 & -0.0474073540 & -0.0474073533 \end{array} \right)$$

Edit

I suppose that you will not need too much terms since it is an alternating series. Checking for $\beta=1$ and writing $$I(1)=\sum_{n=0}^p (-1)^n 2^{2n+1}\frac{E_{2 n+1}(-1)+1}{(2n+1)^2\,(2n)!}+\sum_{n=p+1}^\infty (-1)^n 2^{2n+1}\frac{E_{2 n+1}(-1)+1}{(2n+1)^2\,(2n)!}$$ $$R_p=\Bigg|\frac{2^{2 p+3} (E_{2 p+3}(-1)+1)}{(2 p+3)^2 \Gamma (2 p+3)}\Bigg|$$

Computed for $10 \leq p \leq 100$, a quick and dirty regression $$\log(R_p)=a+b\, p$$ gives (with $R^2=0.999994$) $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -4.54337 & 0.03703 & \{-4.61696,-4.46978\} \\ b & -0.92421 & 0.00061 & \{-0.92541,-0.92300\} \\ \end{array}$$

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  • $\begingroup$ Very nice approach, sir @ClaudeLeibovici! This is clearer now, but do you think there is a closed form to the original problem? Any better approach than mine? $\endgroup$ – lpb Dec 24 '20 at 7:22
  • $\begingroup$ Your steps are satisfactory to me for sure! I just thought my approach was making it difficult to find a closed shape, that's all. Merry Christmas to you and a happy new year! lol $\endgroup$ – lpb Dec 24 '20 at 7:35
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    $\begingroup$ @ClaudeLeibovici Notice that as $\beta$ increases $I(\beta)$ tends to $\pi/2$. I have a mirabilis demonstratio but this space is to little... $\endgroup$ – Raffaele Dec 24 '20 at 13:20

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