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How do you go about integrating this and can it even be done with elementary functions? $$ \int \sin(x)+\sum^{\infty}_{n=1} (-1)^n\frac{\sin^{(2n+1)}(x)}{(2n+1)!} dx. $$

I understand the concept of $$ \int \sum^{\infty}_{n=1}\sin^n (x) dx =\int \frac{dx}{1-\sin x}$$ but I have no clue if that will even help but its the first idea I had to try and clean it up or even how to use it to clean it up.

Additionally this is very similar to integrating $ \int \sin (\sin (x)) $ since the sum is so close to the Taylor series summation of $\sin (x)$ the only difference being $\sin (x)$ is just $x$, not sure if that would help but I'm trying to add as much as I know.

Xander Henderson
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Maxwell Mogadam
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1 Answers1

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$$\sum^{\infty}_{n=1} (-1)^n\frac{t^{(2n+1)}}{(2n+1)!}= \sin (t)-t$$

So, as you noticed, the problem is "simply" to compute $$I=\int \sin(\sin(x))\,dx$$ which cannot be obtained aven using special functions.

If it was for definite integrals, the problem would be different since $$\int_0^{\frac \pi 2} \sin(\sin(x))\,dx=\frac{\pi }{2}\pmb{H}_0(1)$$ where appears the Struve function.

But, from the answer to this question, we have, using Fourier series, $$\color{blue}{\sin(\sin(x)) = 2\sum_{k=0}^\infty J_{2k+1}(1)\sin \left((2k+1)x\right)}$$ which makes $$I=-2\sum_{k=0}^\infty \frac{ J_{2 k+1}(1) }{2 k+1}\cos ((2 k+1) x)$$ and the coefficients vary so fast that very few terms would be sufficient for a definite integral.

For example, using only three terms and integrating from $0$ to $\frac \pi 2$ would give $0.8932437398$ while the exact value given above is $0.8932437410$.

Claude Leibovici
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