Evaluating $\int e^{\sin x}\ dx$ in terms of a Taylor series.

Question

NOW before anyone gets too hasty in saying "oh, it's impossible," I would like to presume that it is possible despite the fact that it obviously is not, and theoretically search for some way of simply representing the solution. I came up with the idea of switching $e^{\sin x}$ into its taylor series and integrating every term. Now I have a new series, but how do I "de-Taylor" this series into some combination of elementary functions? Is this even possible with $e^{\sin x}$? Even if it isn't, is there a method for being able to de-Taylor a series? My equations: $$ \int e^{\sin x}\ dx = x+\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{40}-\frac{x^6}{90}-\frac{x^7}{1680}\dots $$ Surely there is some algorithm that satisfies those denominators, right? There just has to be! And of course, I know how naive this may seem to pursue, but I have taken an interest in this problem and am actively seeking a solution.

Answer 1

Suppose we agree that the solution to our integral is of the following form:

$$\int e^{\sin(x)}dx=f(x)e^{\sin(x)}+C$$

For some $f(x)$. Differentiating both sides reveals that

$$e^{\sin(x)}=[f'(x)+f(x)\cos(x)]e^{\sin(x)}$$

Which becomes equivalent to solving

$$1=f'(x)+f(x)\cos(x)$$

Furthermore, we can assert that $f(x)$ is not an algebraic function, since the LHS of this is algebraic while the RHS would end up being transcendental.

Indeed, the above differential equation was not able to be solved by WolframAlpha, so I highly doubt of a closed form.

Answer 2

There is no such thing as "de-Taylor". This integral can't be combined of elementary functions. Believe me because it's one of the simplest and most famous integrals that can't be combined of elementary functions. In fact only a small minority of integrals and taylor series have a presentation with elementary functions.

Answer 3

To get the power series for $e^{\sin x}$, you can do this:

Let $f(x) = e^{\sin x}$. Then $f'(x) = \cos x\,f(x) $.

Therefore, if $f(x) =\sum_{n=0}^{\infty}a_n x^n $, then $f'(x) =\sum_{n=1}^{\infty}n a_n x^{n-1} =\sum_{n=0}^{\infty}(n+1) a_{n+1} x^{n} $.

Since $\cos(x) =\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!} $,

$\begin{array}\\ f'(x) &=\cos(x)f(x)\\ &=(\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!})(\sum_{n=0}^{\infty}a_n x^n)\\ &=(\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!})(\sum_{n=0}^{\infty}a_n x^n)\\ &=\sum_{m=0}^{\infty} x^m \sum_{k=0}^{\lfloor m/2 \rfloor} (-1)^k \frac{1}{(2k)!}a_{m-2k}\\ &=\sum_{m=0}^{\infty} x^m \sum_{k=0}^{\lfloor m/2 \rfloor} (-1)^k \frac{a_{m-2k}}{(2k)!}\\ \end{array} $

Therefore $(n+1) a_{n+1} =\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \frac{a_{n-2k}}{(2k)!} $.

Since $f(0) = 1$, $a_0 = 1$.

Then, with $n=0$, $a_1 =1 $; with $n=1$, $2a_2 =a_1 =1 $ so $a_2 = \frac12$; with $n=2$, $3a_3 =a_2-a_0/2 =\frac12-\frac12 =0 $; with $n=3$, $4a_4 =a_3-a_1/2 =0-\frac12 =-\frac12 $ so $a_4 = -\frac18 $.

So $f(x) =1+x+\frac{x^2}{2}-\frac{x^4}{8}+... $. Integrating, $\int f(x) =x+\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{40}+... $.

This agrees with Wolfy (there is an error in your coefficient of $x^5$ - you divided by 4 instead of 5).

Note that method (which is not original by me) can be used to get the power series for $e^{g(x)}$ where $g(0) = 0$: let $f(x) = e^{g(x)}$. Then $f'(x) = g'(x)e^{g(x)} =g'(x)f(x) $. Substituting power series for $f$ and $g$, we get a recursion for the coefficients of $g$ in terms of the coefficients of $f$.

Read generatingfunctionology (it's free!) at https://www.math.upenn.edu/~wilf/DownldGF.html for lots more neat stuff.