How is this integral from $0\to\infty$ defined?

Question

I have the following expression $$\int_0^{\infty} \frac{x^2}{(1+x^2)^2}dx$$ and I understand that this can be evaluated to $$\bigg[\frac{1}{2}\bigg(\tan^{-1}(x) - \frac{x}{1+x^2}\bigg)\bigg]_0^\infty$$Now with $\tan^{-1}(x)$ when $x=\infty$ it obviously equals $\frac{\pi}{2}$ however to my mind $\frac{x}{1+x^2}$ is undefined and so this expression shouldn't be defined. However everywhere I've read $($incl. Wolfram alpha$)$ lists the answer as $\frac{\pi}{4}$ suggesting that the second part of the expression is $0$. Can anyone help me with this?

Answer 1

Because $\lim\limits_{x\rightarrow+\infty}\frac{x}{x^2+1}=0$.

The answer is $\frac{\pi}{4}$ of course.

I think the substitution $x=\tan{t}$ gives an easy solution.

Answer 2

@CooperCape Wolfram|Alpha says that your search is undefined because finding a limit as you approach $\infty$ is not the same thing as substituting $\infty$ directly into an equation. Limits have a specific meaning, where everything's nicely defined. But $\infty$ is different since it's not nicely defined. Imagine if you were trying to find $\frac{x}{x}$ when $x$ is really big. Obviously, $\frac{x}{x}=1$ and our limit would tell us this too. But if $2\times\infty=\infty$ then wouldn't $\frac{x}{x}=\frac{\infty}{\infty}=\frac{2\infty}{\infty}=2$ ? But by the same logic, wouldn't it equal $3$ and $4$ and any other number? The point I'm trying to make is that you can't replace $\lim_{x\to\infty}f(x)$ with $f(\infty)$ because, for most cases, the second expression is meaningless.

You'd want to search your question as a limit, which Wolfram|Alpha likes and can work with: https://www.wolframalpha.com/input/?i=lim+x%2F(1%2Bx%5E2)+as+x+to+infinity