Finding a dominating function to evaluate arctan integral

Question

I want to find a dominating function to evaluate the limit of the integral

$$ \int_{-\infty}^{\infty} \arctan\left(\frac{1}{n^2}(b - x)\right) - \arctan\left(\frac{1}{n^2}(a - x)\right) \; dx $$

as $n$ goes to infinity.

Answer 1

The integral appears to be independent of $n.$ Suppose $a<b.$ Then the integral equals

$$\int_{-\infty}^\infty \int_{(a-x)/n^2}^{(b-x)/n^2} \frac{1}{1+t^2}\,dt\, dx = \int_{-\infty}^\infty \int_{a-n^2t}^{b-n^2t}\frac{1}{1+t^2}\, dx\,dt$$ $$ = \int_{-\infty}^\infty (b-a)\frac{1}{1+t^2}\,dt = (b-a)\cdot \pi.$$

I've used Tonelli here. This shows there is no useful dominating function to calculate the limit using the DCT. (If there were, then the limit of the integrals would be $0$ by the DCT.)

Answer 2

Let $a<b$. Let $f_n(t)=\arctan\left(\frac{1}{n^2}(t - x)\right)$. Clearly $f(t)$ is differentiable in $[a,b]$ and $$ f_n'(t)=\frac{1}{n^2+(t-x)^2}\le\frac{1}{1+(x-c)^2}. $$ Hence by the Lagrange Mean Value Theorem, $$ f_n(b)-f_n(a)=f_n'(c)(b-a), c\in(a,b) $$ we have $$ \bigg|\arctan\left(\frac{1}{n^2}(b - x)\right)-\arctan\left(\frac{1}{n^2}(a - x)\right)\bigg|=\frac{1}{n^2+(c-x)^2}(b-a)\le\frac{b-a}{1+(x-c)^2}. $$