Antiderivative of : $t \mapsto (1-t^2)^\lambda$

Question

I would like to find an antiderivative of the function

$$t \mapsto (1-t^2)^\lambda$$ where $\lambda \in \mathbb{R}_{>0}$

I really don't know how to proceed. One idea is to use the generalized binomial theorem to get :

$$(1-t^2)^\lambda = \sum_{k = 0}^{\infty} \binom{\lambda}{k}(-1)^kt^{2k}$$

And by termwise integration I get that a possible antiderivative is

$$\sum_{k = 0}^{\infty} \binom{\lambda}{k}(-1)^k\frac{t^{2k+1}}{2k+1}$$

The problem is that this form isn't really helpful. So is there a close form of this? So that I can study the behavior of the function when $\lambda \to \infty$ on $[0,1]$, for example.

Answer 1

Let $t=\sin \theta$ and substitute to get

$$\int (\cos \theta)^{2\lambda+1} \; d\theta.$$

This is still not a pretty anti-derivative (especially if $\lambda$ is not an integer) but you know exactly what the graph looks like , so you can study the behavior as $\lambda\to\infty$, perhaps.

Answer 2

It turns out that no elementary function is the antiderivative you are looking for. The simplest answer makes use of the hypergeometric function: $$ f(t) = {}_2F_ 1\left(\frac12, -λ, \frac32, -t^2\right) \cdot t + C $$ The differential equation satisfied by ${}_2F_1$ will then give $f'(t) = (1-t^2)^\lambda$.

You can then get a series around $t = 0$ to study the effect of $\lambda$. Intuitively, you'd expect $f(t)$ to go to infinity as $\lambda \to \infty$ for every non-zero $t$.