Arc length of $x\sin x$

Question

Can anyone tell me if it is possible to find a formula for the arc length of the function $f(x)= x\sin x$? I've been working on this on and off for a few days and I'm starting to think it's impossible, yet it seems like such a simple expression.

Wolfram Alpha can give me a determinate answer, but this won't work for me.

This is the integral that should solve for the arc length:

$$ \int \sqrt{ (x \cos(x)+\sin(x))^2 + 1\,}\,\, \mathrm{d}x $$

I can't seem to find a solution for it. Can anyone help me, or explain why it can't be solved?

Answer 1

In principle, you can use the Risch algorithm to prove that $\sqrt{(x\cos(x)+\sin(x))^2+1}$ has no elementary antiderivative.
However, this is the mixed algebraic-transcendental case, which is quite difficult: computer algebra systems such as Maple don't implement it fully.

As for non-elementary special functions, that's pretty much wide open. You could always define a new one, "Nick's function"...

As for the "why", the fact is that, outside of certain rather special types of function for which standard techniques work, "most" elementary functions do not have closed-form antiderivatives. This may come as a surprise to survivors of calculus courses, where almost every question does have a closed-form solution.

Answer 2

Robert Israel gave the answer to the problem.

But, let us consider the case where you need to get a number for the arc length between $x=0$ and $x=a$. What you can do is to expand the integrand as a Taylor series around $x=0$ and use as many terms as you can handle. For example $$\sqrt{ (x \cos(x)+\sin(x))^2 + 1\,}=1+2 x^2-\frac{10 x^4}{3}+\frac{629 x^6}{90}-\frac{12329 x^8}{630}+\frac{1770127 x^{10}}{28350}+O\left(x^{11}\right)$$ Integrating between $0$ and $a$ gives for the arc length $$L=a+\frac{2 a^3}{3}-\frac{2 a^5}{3}+\frac{629 a^7}{630}-\frac{12329 a^9}{5670}+\frac{1770127 a^{11}}{311850}+\cdots$$ For sure, this will only work for small values of $a$. Just for your curiosity, I give below a few results $$\left( \begin{array}{ccc} a & approx & "exact" \\ 0.1 & 0.10066 & 0.10066 \\ 0.2 & 0.205132 & 0.205132 \\ 0.3 & 0.316566 & 0.316564 \\ 0.4 & 0.437144 & 0.437070 \\ 0.5 & 0.568825 & 0.567681 \\ 0.6 & 0.718789 & 0.708479 \\ 0.7 & 0.923335 & 0.858767 \end{array} \right)$$

If you need to work with larger values of $a$, you should need to consider numerical integration.

Edit *(five years later)

Revisitng this answer, I found that the initial series expansion is quite bad and that much better results could be obtained using the simplest Padé approximant $$P_2=\frac{11 x^2+3}{5 x^2+3}$$ But, surprising is the fact that the next one $$P_4=\frac{2184053 x^4+1223400 x^2+162540}{929213 x^4+898320 x^2+162540}$$ is worse that $P_2$.

Then, came the idea of finding a rational function which could be better than both of them. The following approximation is $$Q=\frac{-\frac{1943 x^4}{1178}+\frac{2208 x^2}{755}+1}{-\frac{365 x^4}{594}+\frac{986 x^2}{1039}+1}$$ which is simple to integrate.

Just to compare the infinite norms $$\int_0^1 \Big[\sqrt{ 1+(x \cos(x)+\sin(x))^2 }-P_2\Big]^2=1.07\times 10^{-4}$$ $$\int_0^1 \Big[\sqrt{ 1+(x \cos(x)+\sin(x))^2 }-P_4\Big]^2=6.56\times 10^{-4}$$ $$\int_0^1 \Big[\sqrt{ 1+(x \cos(x)+\sin(x))^2 }-Q\Big]^2=8.41\times 10^{-8}$$

$$\left( \begin{array}{ccc} a & \text{approximation} & \text{exact} \\ 0.1 & 0.100653 & 0.100660 \\ 0.2 & 0.205089 & 0.205132 \\ 0.3 & 0.316479 & 0.316564 \\ 0.4 & 0.436979 & 0.437070 \\ 0.5 & 0.567623 & 0.567681 \\ 0.6 & 0.708449 & 0.708479 \\ 0.7 & 0.858726 & 0.858767 \\ 0.8 & 1.017170 & 1.017250 \\ 0.9 & 1.182080 & 1.182140 \\ 1.0 & 1.351280 & 1.351340 \end{array} \right)$$