Check for uniform convergence of the integral $\int\limits_{0}^{1}\frac{\arctan ax}{(1-x^2)^a}dx$

Question

Check for uniform convergence of the integral $$\int\limits_{0}^{1}\frac{\arctan ax}{(1-x^2)^a}dx, \; a\in (-1,1)$$ My attempt was an evaluation, but I don't think it's right

$$\left| \int\limits_{0}^{1}\frac{\arctan ax}{(1-x^2)^a}dx \right| \leq \frac{\pi}{2} \int\limits_{0}^{1}\frac{1}{(1-x^2)^a}dx \leq \frac{\pi}{2} \int\limits_{0}^{1}\frac{1}{(1-x)^a}dx < \infty$$

Answer 1

I think I can establish the estimate.

My first issue was, as I mentioned, the inequality from the center to the right. With $x \in (0, 1)$ we can establish that

$$ x^2 \leq x \implies \frac{1}{1-x^2} \leq \frac{1}{1-x}.$$

For the moment, I'll omit the justification (it's an exercise in inequality manipulations). Now, you already used the fact that $\arctan x \leq \frac{\pi}{2}$ for each $x$, so it remains to handle the right inequality. Observe that by the change of variable $x \mapsto 1-x$ the integral remains unchanged. Now,

\begin{align*} \int_{0}^{1}\frac{1}{x^a} \ dx &= \frac{x^{-a+1}}{1-a}\Bigg|_0 ^1 \\ &= \frac{1}{1-a} \\ & \leq \frac{1}{2}. \\\ \end{align*}

Observe that the inequality in the last line is achieved by allowing $a \to -1^+$.

Upon writing this I've noticed that if $a \to 1^-$ then the integral can be made arbitrarily large. This may pose a problem - so I think it's safe to say that the original integral is uniformly convergent on $(-1, 1-\epsilon)$ where $\epsilon > 0.$ In fact, the left boundary can be allowed to go to $-\infty.$

TL:DR; So long as $a$ is an interval far enough away from $1$, say, for example $[-2, \frac{2}{3}]$, we have uniform convergence. No extra work needs to be shown for this.

Answer 2

If the integral converges uniformly, then (Cauchy criterion) for every $\epsilon > 0$ there exists $\delta \in (0,1)$ such that for every $c_1,c_2$ such that $\delta < c_1 < c_2 < 1$ and every $a \in (-1,1)$ we would have

$$\tag{*}\left|\int_{c_1}^{c_2} \frac{\arctan ax}{(1-x^2)^a} \, dx \right| < \epsilon$$

For any such $\delta$, we can take $c_1 = (1-\frac{1}{n}) $, $c_2 = (1-\frac{1}{2n}) $ and $a_n = (1-\frac{1}{n}) $ and for all sufficiently large $n$ we have $c_1,c_2 \in (\delta ,1)$ and $a_n \in (-1,1)$. It follows that

$$\left|\int_{c_1}^{c_2} \frac{\arctan a_nx}{(1-x^2)^{a_n}} \, dx \right| = \int_{1-\frac{1}{n}}^{1 - \frac{1}{2n}} \frac{\arctan [(1-\frac{1}{n})x]}{(1-x^2)^{1 - \frac{1}{n}}} \, dx> \frac{1}{2n} \cdot\frac{\arctan [(1-\frac{1}{n})(1-\frac{1}{n})]}{\left(1-(1-\frac{1}{2n})^2\right)^{1-\frac{1}{n}}}\\ = \frac{1}{2}\frac{\arctan [(1-\frac{1}{n})^2]}{n\left(\frac{1}{n}- \frac{1}{4n^2}\right)^{1 - \frac{1}{n}}}= \frac{1}{2}\frac{\arctan [(1-\frac{1}{n})^2]}{\frac{n}{n^{1 - \frac{1}{n}}}\left(1- \frac{1}{4n}\right)^{1 - \frac{1}{n}}}= \frac{1}{2}\frac{\arctan [(1-\frac{1}{n})^2]}{n^{\frac{1}{n}}\left(1- \frac{1}{4n}\right)^{1 - \frac{1}{n}}}$$

The RHS converges to $\frac{\pi}{8}$ as $n \to \infty$ and must exceed $\frac{\pi}{16}$ for all sufficiently large $n$. This implies that the condition (*) for uniform convergence is violated when we take small values of $\epsilon > 0$.

Thus, the integral fails to converge uniformly for all $a \in (-1,1)$ with the problem arising because values of $a$ can be arbitrarily close to $1$.

Answer 3

If you enjoy generalized hypergeometric functions, there is an explicit solution

$$I(a)=\int\limits_{0}^{1}\frac{\tan^{-1} (ax)}{(1-x^2)^a}\, dx \qquad\qquad \text{with} \qquad a <1$$

Using the series expansion of the numerator, we then have $$\sum_{n=0}^\infty (-1)^n\,\frac{a^{2 n+1} }{2 n+1}\,\frac{ x^{2 n+1}}{ \left(1-x^2\right)^{a} }$$ $$\int_0^1 \frac{ x^{2 n+1}}{ \left(1-x^2\right)^{a} }\,dx=\frac{\Gamma (1-a) \,\,\Gamma (n+1)}{2\, \Gamma (n+2-a)}$$ Summing and simplifying $$\large\color{blue}{I(a)=\frac{a}{2 (1-a)}\,\,\, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},2-a;-a^2\right)}$$

Expanding around $a=0$

$$I(a)=\frac{a}{2 (1-a)}\left(1-\frac{a^2}{6}\sum_{n=0}^\infty \alpha_n \,a^n\right)$$ where the first coefficients are $$\left\{1,\frac{1}{2},-\frac{3}{20},-\frac{5}{24},\frac{331}{5040},\frac {865}{6048},-\frac{2123}{181440},-\frac{61399}{680400},-\frac{94 2617}{513216000}\right\} $$ Notice that $$\sum_{n=0}^\infty \alpha_n \,a^n=\frac{1+\frac{409 }{798}a+\frac{184 }{665}a^2 }{1+\frac{5 }{399}a+\frac{671 }{1596}a^2 }$$ whose error is $\sim \frac{a^5}{18}$.

This should allow to generate extremely sharp bounds. Function $I(a)$ goes through a minimum value around $a=-2.95$ and tends to $0^-$ when $a$ tends to $-\infty$.

Edit

If $a\to 1^-$, the asymptotic is $$I(a)=\frac{a \, \pi}{8 (1-a)} +O\left((a-1)\right)$$