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

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.

• See here. Commented Nov 26, 2016 at 0:19

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.

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.

• This doesn't appear to be what the OP wants. Commented Nov 26, 2016 at 0:14
• Yeah, this successfully shows how I got where I am. Thanks for that, but How can I now express that series as a new function, even just a generalized summation? Commented Nov 26, 2016 at 0:17
• @dsillman2000 I'm looking for that algorithm. Its somewhere XD Commented Nov 26, 2016 at 0:18
• It's just a bunch of coefficients. I strongly doubt that there is a general form for them. Commented Nov 26, 2016 at 0:19

I have a slightly simplify series form of that:

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

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

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

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,

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

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

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

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

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

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

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

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

If you still insist on its Taylor-series form, you can have this procedure:

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

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

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

$$=\int\sum\limits_{n=0}^\infty\dfrac{C_n^{2n}}{4^n(2n)!}~dx+\int\sum\limits_{n=1}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^mC_{n+m}^{2n}\cos2mx}{2^{2n-1}(2n)!}~dx+\int\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\dfrac{(-1)^mC_{n+m+1}^{2n+1}\sin((2m+1)x)}{4^n(2n+1)!}~dx$$

$$=\sum\limits_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^m\sin2mx}{4^nm(n+m)!(n-m)!}-\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\dfrac{(-1)^m\cos((2m+1)x)}{4^n(n+m+1)!(n-m)!(2m+1)}+C$$

$$=\sum\limits_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{m=1}^\infty\sum\limits_{n=m}^\infty\dfrac{(-1)^m\sin2mx}{4^nm(n+m)!(n-m)!}-\sum\limits_{m=0}^\infty\sum\limits_{n=m}^\infty\dfrac{(-1)^m\cos((2m+1)x)}{4^n(n+m+1)!(n-m)!(2m+1)}+C$$

$$=\sum\limits_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^m\sin2mx}{4^{m+n}m(2m+n)!n!}-\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^m\cos((2m+1)x)}{4^{m+n}(2m+n+1)!n!(2m+1)}+C$$

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

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

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.

• This is not true. You can set up complicated DE and attempt to solve. Commented Nov 25, 2016 at 23:54
• Did you contribute to the -3 rating of my answer because you thought I am wrong? And in your own answer you admit that I am right? I find this extremely unfair. Commented Nov 26, 2016 at 10:53
• No where in my answer do I propose that it is impossible to undo a Taylor series. Indeed, I downvoted because this question shows a lack of effort and is also not true. How do you define "only a small minority of integral and taylor series have a [representation] with elementary functions"? And even if true, it is like saying "most numbers are irrational, so the answer is most likely false". Clearly, this needs improvement. Commented Nov 26, 2016 at 12:28