Limit and integral

**chiph588@** · February 25th 2011, 02:38 PM

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

**girdav** · February 27th 2011, 01:31 PM

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
$\displaystyle I_n = \int_0^{+\infty}\dfrac{\tan^2 x}{(n+\tan^2 x)(1+x^2)}dx<br /> =\sum_{j=0}^{+\infty}\int_{j\pi}^{(j+1)\pi}\dfrac{ \tan^2 x}{(n+\tan^2 x)(1+x^2)}dx <br />$
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 $0$ .

**Bruno J.** · February 27th 2011, 06:43 PM

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

**Random Variable** · February 27th 2011, 06:53 PM

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?

**Bruno J.** · February 27th 2011, 07:14 PM

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

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