Does an elementary antiderivative of $e^{\sin x} \sin x$ exist?

Question

I wonder if an elementary antiderivative of the function $e^{\sin x} \sin x$ exist? If so, could anyone help me to derive this certain antiderivative step by step? If not, is a strict proof of the nonexistence available, maybe by using knowledge of differential algebra (with which I'm not familiar)? Thanks in advance!

Answer 1

Write $$e^{\sin (x)}\, \sin (x)=\sum_{n=0}^\infty \frac{\sin ^{n+1}(x)}{n!}$$ $$\int e^{\sin (x)}\, \sin (x)\,dx=\sum_{n=0}^\infty \frac 1{n!}\int \sin ^{n+1}(x)\,dx$$ Use the reduction formula $$I_n=\int \sin ^{n}(x)\,dx \quad \implies \quad I_n= \dfrac {n - 1} n I_{n - 2} - \dfrac {\sin^{n - 1}( x) \cos (x)} n $$ and, for definite integrals, you can generate nice identities such as $$\int_0^{\frac \pi 2} e^{\sin (x)}\, \sin (x)\,dx=\sum_{n=0}^\infty \frac 1{n!} \frac{\sqrt{\pi } \,\,\Gamma \left(\frac{n}{2}+1\right)}{2\, \Gamma \left(\frac{n+3}{2}\right)}=\frac{\pi}{2} (\pmb{L}_{-1}(1)+I_1(1))$$ where appear Struve and Bessel functions.