# Math Help - Limit and integral

1. ## Limit and integral

Compute $\displaystyle \lim_{n\to\infty} \int_0^\infty \frac{\tan^2x}{(n+\tan^2x)(x^2+1)}dx$.

2. Let $\displaystyle I_n:= \int_0^{+\infty}\dfrac{\tan^2 x}{(n+\tan^2 x)(1+x^2)}dx$. The sequence $\left\{I_n\right\}_n$ is decreasing and positive so the limit exists. We have
$\displaystyle I_n = \int_0^{+\infty}\dfrac{\tan^2 x}{(n+\tan^2 x)(1+x^2)}dx
=\sum_{j=0}^{+\infty}\int_{j\pi}^{(j+1)\pi}\dfrac{ \tan^2 x}{(n+\tan^2 x)(1+x^2)}dx
$

and by the substitution $t=x-j\pi$ we get $\displaystyle I_n =\sum_{j=0}^{+\infty}\int_0^{\pi}\dfrac{\tan^2 t}{(n+\tan^2t)(1+(t+j\pi)^2)}dt$.
For all $j$ we have the inequality $\dfrac 1{1+((j+1)\pi)^2}\leq \dfrac 1{1+(t+j\pi)^2}\leq \dfrac 1{1+(j\pi)^2}$ hence $\displaystyle\int_0^{\pi}\dfrac{\tan^2 t}{n+\tan^2t}dt \sum_{j=0}^{+\infty}\dfrac 1{1+((j+1)\pi)^2}\leq I_n\leq \int_0^{\pi}\dfrac{\tan^2 t}{n+\tan^2t}dt \sum_{j=0}^{+\infty}\dfrac 1{1+(j\pi)^2}$.
Now we try to compute $\displaystyle J_n :=\int_0^{\pi}\dfrac{\tan^2 t}{n+\tan^2t}dt = \int_0^{\frac{\pi}2}\dfrac{\tan^2 t}{n+\tan^2t}dt +\int_0^{\frac{\pi}2}\dfrac 1{n\tan^2t+1}dt$. By putting $x=\tan t$ we get
$\displaystyle J_n =\int_0^{+\infty}\dfrac{x^2}{(n+x^2)(1+x^2)}dx+\in t_0^{+\infty}\dfrac{1}{(nx^2+1)(1+x^2)}dx$. We can see that the two integrals are the same (by the substitution $y=\frac 1x$). So we only have to compute the first one.
We have $\displaystyle \dfrac{x^2}{(n+x^2)(1+x^2)}dx =\dfrac 1{n-1}\left(\dfrac n{n+x^2}-\dfrac 1{1+x^2}\right)$. Then $\displaystyle J_n =2\dfrac 1{n-1}\int_0^{+\infty}\dfrac {dt}{1+\left(\dfrac t{\sqrt n}\right)^2}-2\dfrac{\pi}{2(n-1)} =\dfrac{(\sqrt n-1)\pi}{n-1}$ and we conclude the limit we were looking for is $0$.

3. It is an immediate consequence of the dominated convergence theorem that the limit is 0.

4. Originally Posted by Bruno J.
It is an immediate consequence of the dominated convergence theorem that the limit is 0.
Would you mind showing why the DCT applies?

5. The integrand is non-negative and converges monotonically to $0$ almost everywhere, so you may switch the limit and the integral.