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I was curious about this integral so I asked about its convergence here: Does $\int_{-\infty}^{\infty}\sin^{-1}\left(\frac{\sin(x)}{x}\right)dx$ converge?

@jizert answered, explaining that it did converge. My follow-up question is what it converges to?

I pasted my work from the the other question below:

My attempts: $$\frac{dy}{dx}=\sin^{-1}\left(\frac{\sin(x)}{x}\right)$$ $$\sin\left(\frac{dy}{dx}\right)=\frac{\sin(x)}{x}$$ $$x\sin\left(\frac{dy}{dx}\right)=\sin(x)$$ Seems like a dead end. $$\int_{-\infty}^{\infty}\sin^{-1}\left(\frac{\sin(x)}{x}\right)dx$$ $$2\int_{0}^{\infty}\sin^{-1}\left(\frac{\sin(x)}{x}\right)dx$$ $$2\int_{1}^{0}\sin^{-1}(u)\frac{du}{\frac{x\cos(x)-\sin(x)}{x^2}}$$ Also seems like a dead end because I can't figure out how to solve for $x$ in $u=\frac{\sin x}{x}$.

Edit: I also see that $\left|\frac{\sin(x)}{x}\right|\le\left|\sin^{-1}\left(\frac{\sin(x)}{x}\right)\right|\le\left|\frac{\pi}{2}\frac{\sin(x)}{x}\right|$ but I doubt anything can be done with that.

Dylan Levine
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    I see no reason to think there is a closed-form expression for this value. – GEdgar Feb 04 '24 at 01:55
  • Not related to solving but you can say $x=\operatorname{sinc}^{-1}(u)$ – Henry Lee Feb 04 '24 at 20:15
  • Maybe someone else can apply the residue theory and get an infinite series. Also just maybe you can define $I(a)=\int_{-\infty}^\infty\sin(i a\frac{\sin(x)}x),I’(a)=\int_{-\infty}^\infty\frac{i\sin(x)}{|x+ia\sin(x)|}dx$ and solve $u=x+i a\sin(x)$ – Тyma Gaidash Feb 04 '24 at 23:53

1 Answers1

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Using Taylor expansion of $\arcsin{x}$, we have $$\arcsin (x)=\sum_{n=0}^{\infty} \frac{1}{2^{2 n}}\dbinom{2n}{n} \frac{x^{2 n+1}}{2 n+1}$$ and using this result for integration of powers of $\sin(x)/x$
$$\int_{0}^{\infty} \frac{\sin^m(x)}{x^m} dx=\frac{\pi m}{2^m} \sum_{k=0}^{\left\lfloor\frac{m}{2}\right\rfloor} \frac{(-1)^k(m-2 k)^{m-1}}{k !(m-k) !}$$ Combining these two results, we obtain $$ \int_{-\infty}^{\infty} \arcsin{\left( \frac{\sin(x)}{x}\right)}dx = \sum_{n=0}^{\infty} \dbinom{2n}{n} \frac{\pi}{2^{4n}} \sum_{k=0}^{n} \frac{(-1)^k (2(n-k)+1)^{2n}}{k! (2n-k+1)!} \approx 3.96.$$ Edit : Summing up $5000$ terms in the sum, the better approximation is $3.96363$.

Sam
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