What is the integral of $ \int e^x \sin^{-1}xdx $?

Question

How does one integrate: $$ I= \int e^x sin^{-1}xdx $$ I've tried doing this by parts but in the end, I just get I=I. Does this mean the function is "unintegratable"(if something as such exists)?

PS: I'm very new to calculus so this question might just be stupid.

Thanks.

Answer 1

For any $z\in(-1,1)$ we have

$$ \int_{0}^{z}e^{-x}\arcsin(x)\,dx = \int_{0}^{z}e^{-x}\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n}x^{2n+1}\,dx $$ hence the LHS can be represented as a series involving the incomplete $\Gamma$ function.
As an alternative one may exploit the Jacobi-Anger expansion $$ e^{-z\sin\theta}=\sum_{n\in\mathbb{Z}}J_n(iz)\,e^{in\theta}=I_0(z)+2\sum_{n\geq 1}(-1)^n\left[I_{2n-1}(z)\sin((2n-1)\theta)+I_{2n}(z)\cos(2n\theta)\right] $$ to state $$\begin{eqnarray*}\int_{0}^{z}e^{-x}\arcsin(x)\,dx &=& \int_{0}^{\arcsin z}e^{-\sin\theta}\theta\cos\theta\,d\theta\\ &=&I_0(1)\left(-1+\sqrt{1-z^2}+z\arcsin z\right)\\&+&\scriptstyle 2\sum_{n\geq 1}(-1)^n\left[I_{2n-1}(1)\int_{0}^{\arcsin z}\theta\cos\theta\sin((2n-1)\theta)\,d\theta+I_{2n}(1)\int_{0}^{\arcsin z}\theta\cos\theta\cos(2n\theta)\,d\theta\right]\end{eqnarray*}$$ which is a typesetting nightmare with better convergence properties, since $0\leq I_n(1)\leq \frac{I_0(1)}{2^n n!}$.
Here $I_n$ is a modified Bessel function of the first kind.

Answer 2

$$\int e^x\arcsin{x}dx=e^x\arcsin{x}-\int\frac{e^x}{\sqrt{1-x^2}}dx$$ and bye...

Answer 3

$\int e^xf(x)dx =e^xf(x)-\int e^xf'(x)dx =e^xf(x)-e^xf'(x)+\int e^xf''(x)dx =...$

Continue until tired.