Why does $\frac{x^{2}}{1+x^{2}}$ diverge?

Question

Why does $\frac{x^{2}}{1+x^{2}}$ diverge?

I am trying to show that the integral $2\int\int_{\mathbb{R}^{+}}\frac{x^{2}}{(1+x^{2})(1+y^{2})}dxdy$ diverges. The solution says it diverges because $\frac{x^{2}}{1+x^{2}}\geq \frac{1}{2}$. What does this mean? What theorem is being used in analysis?

Answer 1

It's not the function $f(x)=\frac{x^2}{1+x^2}$ that diverges (at $\infty$): indeed its limit at $\infty$ is $1$.

It's the integral $$ \int_0^\infty \frac{x^2}{1+x^2}\,dx $$ that diverges, for the very reason that the limit of the function at $\infty$ is not zero.

Since the limit is $1$, there exists $k>0$ such that, for $x>k$, $f(x)>1/2$. Hence, for $t>k$, $$ \int_0^{t}f(x)\,dx=\int_0^k f(x)\,dx+\int_k^t\frac{1}{2}\,dx= \int_0^k f(x)\,dx+\frac{1}{2}(t-k) $$ and the last term of course diverges for $t\to\infty$.

Answer 2

$\dfrac{x^2}{x^2+1} =1-\dfrac{1}{x^2+1};$

For $x \ge 1$:

$x^2+1 \ge 2$, and

$f(x):= \dfrac{x^2}{x^2+1} \ge 1/2.$

$\displaystyle \int_{1}^{t} (1/2)dx \le \int_{1}^{t}f(x)dx$;.

$(1/2)(t-1) \le \displaystyle \int_{1}^{t}f(x)dx$,

divergent for $t \rightarrow \infty.$

Answer 3

The theorem being used is that if $f(x,y) \geq g(x,y)$, and if $\iint_R g(x,y)\,dx\,dy$ diverges, then $\iint_R f(x,y)\,dx\,dy$ diverges, too.

In your case, since $$\frac{x}{1+x^2} \geq \frac{1}{2}$$ it follows that your integrand satisfies $$f(x,y) = \frac{x^2}{(1+x^2)(1+y^2)} \geq \frac{1}{2}\frac{1}{1+y^2} = g(x,y)$$ Since the integral $$\int_0^\infty \int_0^\infty \frac{1}{2}\frac{1}{1+y^2}\,dx\,dy$$ diverges (check this yourself), it follows that your original integral diverges.