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Let $I(b)$ is the following integral

$$I(b)=\int_0^b \sin(\sin(x)) \, dx.$$

There are some $b$ value for that we know a closed-form of $I(b)$ in term of Struve function $\mathbf{H}_n(x)$. For example

$$\begin{align} I(\pi/2) = & \, \frac{\pi}{2}\mathbf{H}_0(1) \\ I(\pi) = & \, \pi\mathbf{H}_0(1) \\ I(3\pi/2) = & \, \frac{\pi}{2}\mathbf{H}_0(1) \\ I(2\pi) = & \, 0. \end{align}$$

I know that the closed-form of $I(k\pi/2)$ comes from the integral form of Struve function, but perhaps there are other closed forms of this type of integrals.

Question. Is there a closed-form of $I(\pi/4)=\int_0^{\pi/4} \sin(\sin(x)) \, dx$?

user153012
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  • No, there aren't known to be any. Unless, of course, you are willing to accept “incomplete” Struve functions as closed forms. – Lucian Oct 18 '14 at 00:03
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    Conjecture for $I(\pi/4)$ Integral $$I(\pi/4)=\frac{\pi \pmb{H}_0(1)}{2} - \frac{1}{\sqrt{2}} \sum _{n=1}^{\infty } \frac{(-1)^{n-1} \sum _{k=0}^{n-1} \left(-\frac{1}{2}\right)^k \binom{-1/2}{k}}{((2 n-1)\text{!!})^2}$$ – James Arathoon Jan 22 '19 at 04:26
  • +1 @JamesArathoon I think that is correct ... Need time to think. – mick Mar 22 '21 at 23:31
  • Searching “Kampe de Feriet” function may give a closed form – Тyma Gaidash Jun 10 '23 at 22:20

1 Answers1

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$\int_0^b\sin(\sin x)~dx$

$=\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx$

$=-\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}x}{(2n+1)!}d(\cos x)$

$=\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2x)^n}{(2n+1)!}d(\cos x)$

$=\int_0^b\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}x}{(2n+1)!}d(\cos x)$

$=\int_0^b\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}x}{(2n+1)!k!(n-k)!}d(\cos x)$

$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}\right]_0^b$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!(\cos^{2k+1}b-1)}{(2n+1)!k!(n-k)!(2k+1)}$

Harry Peter
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