How to integrate $\sqrt{\arctan(x)}$

Question

How to do $$\int\sqrt{\arctan(x)}\, \mathrm dx \:??? $$ Is there any other special function defined like this?

Answer 1

I am not sure about a closed form, but this one can be solved rather well using a series expansion. In particular, using the substitution

$$u = \arctan(x),\ dx = \sec^2 u\ du$$

we get

$$\int \sqrt{\arctan(x)} dx = \int \sqrt{u} \sec^2 u\ du$$

as was remarked in the comments. However, now the trick is to form the right series expansion of the integrand. Directly expanding the two factors in terms of Taylor series is actually not the best approach because it quickly becomes rather messy. Instead, it turns out that the best is to actually expand $\sec^2 u$ first. We can do this by noting that it is the derivative of $\tan u$ and then using the series for that, which is

$$\tan{u} = \sum_{n=0}^{\infty} \frac{U_{2n+1}}{(2n+1)!} {u}^{2n+1}$$

where $U_n$ are the so-called "up/down numbers", given recursively by

$$U_0 := 1,\ U_{n+1} := \frac{1}{2} \sum_{k=0}^{n} \binom{n}{k} U_k U_{n-k}$$

and so given that $\frac{d}{du} \tan u = \sec^2 u$ we get

$$\sec^2 u = \sum_{n=0}^{\infty} \frac{U_{2n+1}}{(2n+1)!} (2n+1) u^{2n} = \sum_{n=0}^{\infty} \frac{U_{2n+1}}{(2n)!} u^{2n}$$

hence using that $\sqrt{u} = u^{1/2}$ we get the "Puiseux series"

$$\sqrt{u} \sec^2 u = \sum_{n=0}^{\infty} \frac{U_{2n+1}}{(2n)!} u^{2n\ +\ 1/2}$$

and then we can integrate

$$\int \sqrt{u} \sec^2 u\ du = \sum_{n=0}^{\infty} \frac{U_{2n+1}}{(2n)!} \frac{u^{2n\ +\ 3/2}}{\left(2n + \frac{3}{2}\right)}$$

hence

$$\int \sqrt{\arctan(x)} dx = C + \sum_{n=0}^{\infty} \frac{U_{2n+1}}{(2n)!} \frac{[\arctan(x)]^{2n\ +\ 3/2}}{\left(2n + \frac{3}{2}\right)}$$

Since the original power series for $\tan u$ converges in all $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, and $\arctan(x)$ has this as its range, I believe the above series should be convergent for all $x \ge 0$, which is also the domain of the original integrand, hence providing a complete description of the antiderivative.

Now there might be some way to make this into some kind of hypergeometric function, because apparently we have a weird formula

$$\tan x = \frac{8x}{\pi^2 - 4x^2}\ _3F_2\left(1, \frac{1}{2} - \frac{x}{\pi}, \frac{x}{\pi} + \frac{1}{2}; \frac{3}{2} - \frac{x}{\pi}, \frac{x}{\pi} + \frac{3}{2}; 1\right)$$

from

http://functions.wolfram.com/ElementaryFunctions/Tan/26/01/

but I have no idea how this is obtained - note that $x$ is not the series variable for the hypergeometric series, so this likely requires something rather clever. But it might also be applicable to the above series as well in some way. I note that Wolfram's integrator doesn't seem to be able to give more peculiar forms like this where the variable of integration appears outside its usual place as in the series variable, i.e. the last argument to the hypergeometric function.

Answer 2

Try this: $ x = \tan y $. Thus, $ \arctan(x) = y $ and $ dx = \sec^2 y dy $.

$$ \int\sqrt{\arctan(x)}dx = \int\sqrt{y} \sec^2(y) dy $$