Does an elementary indefinite integral for 2^sin(x) exist?

Question

Wolframalpha says no, and I've had little luck elsewhere, maybe because search engines aren't the best at making sense of the "^" symbol.

And if not, are any numerical methods known for solving this integral?

Answer 1

There is no known closed form I am aware of, but it may be worth noting that

$$J_0(-i\ln(2))=\int_a^{a+2\pi}2^{\sin(x)}~\mathrm dx$$

where $J$ is a Bessel function. As far as numerical goes you can use any standard method.

Answer 2

Notice that: $$2^{\sin(x)}=\left(e^{ln(2)}\right)^{\sin(x)}=e^{\ln(2)\sin(x)}$$ And you will find articles like this that show this is non-elementary.

You can represent it as a series of integrals by using the fact that: $$e^z=\sum_{n=0}^\infty\frac{z^n}{n!},z\in\mathbb{C}$$

Answer 3

As Henry Lee answered $$2^{\sin(x)}=e^{\ln(2)\sin(x)}=\sum_{n=0}^\infty \frac{\log^n(2)}{n!}\sin^n(x)$$ $$\int2^{\sin(x)}\,dx=\sum_{n=0}^\infty \frac{\log^n(2)}{n!}\int\sin^n(x)\,dx$$ and the reduction formula $$\int\sin^n(x)\,dx= -\frac{\cos (x) \sin^{n-1}(x)}{n} + \frac{(n-1)}{n} \int \sin^{n-2} (x) dx$$

For example, considering the partial sums $$S_p(a)=\sum_{n=0}^p \frac{\log^n(2)}{n!}\int_0^{a}\sin^n(x)\,dx$$ as shown below, it converges quite fast $$\left( \begin{array}{ccccc} p & S_p\left(\frac{\pi }{6}\right)& S_p\left(\frac{\pi }{4}\right)&S_p\left(\frac{\pi }{3}\right)&S_p\left(\frac{\pi }{2}\right) \\ 0 & 0.5235987756 & 0.785398163 & 1.047197551 & 1.570796327 \\ 1 & 0.6164628892 & 0.988416272 & 1.393771141 & 2.263943507 \\ 2 & 0.6273434773 & 1.022696374 & 1.467542882 & 2.452616965 \\ 3 & 0.6282952403 & 1.026993004 & 1.479106238 & 2.489619704 \\ 4 & 0.6283616677 & 1.027421244 & 1.480540584 & 2.495285250 \\ 5 & 0.6283655248 & 1.027456676 & 1.480687809 & 2.495996373 \\ 6 & 0.6283657166 & 1.027459182 & 1.480700698 & 2.496071985 \\ 7 & 0.6283657249 & 1.027459337 & 1.480701682 & 2.496078957 \\ 8 & 0.6283657252 & 1.027459346 & 1.480701749 & 2.496079525 \\ 9 & 0.6283657253 & 1.027459346 & 1.480701753 & 2.496079566 \\ 10 & 0.6283657253 & 1.027459346 & 1.480701753 & 2.496079569 \end{array} \right)$$

Answer 4

$\int 2^{\sin x}~dx$

$=\int e^{\ln2\sin x}~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$

$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}\right)~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$

For $n$ is any natural number,

$\int\sin^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts.

For $n$ is any non-negative integer,

$\int\sin^{2n+1}x~dx$

$=-\int\sin^{2n}x~d(\cos x)$

$=-\int(1-\cos^2x)^n~d(\cos x)$

$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$

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

$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}\right)~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$

$=x+\sum\limits_{n=1}^\infty\dfrac{x\ln^{2n}2}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\ln^{2n}2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\ln^{2n+1}2\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{x\ln^{2n}2}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\ln^{2n}2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\ln^{2n+1}2\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$