Why do we split improper integrals where both bounds are at infinity?

Question

I know that when a function $f(x)$ has a discontinuity at some $x = c$, to calculate $\int_{-\infty}^{\infty}f(x)dx$, you need to split the integral into two parts: $\int_{-\infty}^{c}f(x)dx + \int_{c}^{\infty}f(x)dx$.

However, if a function $f(x)$ does not have a discontinuity, why is this split still necessary?

Example: Calculate $\int_{-\infty}^{\infty}x\sin(x^2)dx$.

This integral diverges if we split it at some $x = c$, but if we calculate the integral without splitting it:

$$ \int_{-\infty}^{\infty} x\sin(x^2) dx = \left. -\frac12 \cos(x^2) \right|_{-\infty}^{\infty} \\ = \lim_{x \rightarrow \infty} -\frac12 \cos(x^2) - \lim_{x \rightarrow -\infty} -\frac12 \cos(x^2) \\ = -\frac12(\lim_{x \rightarrow \infty} \cos(x^2) - \lim_{x \rightarrow -\infty} \cos(x^2)) $$

Wouldn't this equal $0$ because $x^2$ would "strip" the negative from negative infinity, making it $-\frac12(\lim_{x \rightarrow \infty} \cos(x^2) - \lim_{x \rightarrow \infty} \cos(x^2))$?

Answer 1

The short answer is that your integral is actually a double limit: $$\lim_{a,b \to \infty} \int_{-a}^b f(x) dx.$$

To show that this limit exists, you have to show that regardless of the speeds at which $a$ and $b$ approach infinity, your integral will approach a single finite value.

What you are proposing to do is: $$\lim_{a \to \infty} \int_{-a}^a f(x) dx.$$

This is the case when both upper and lower limits approach infinity at the same speed which is far less strong than the above kind of integral.

Answer 2

IMO the easiest way to understand this is to consider a very simple integral of this type. Probably the simplest is $$I=\int_{-\infty}^\infty x\,dx\ .$$ Your argument is that the positive bits and the negative bits have the same size and therefore cancel each other out. But both of these sizes are infinite, so this is essentially the same as claiming that $\infty-\infty=0$. Definitely not true! - you cannot do arithmetic with $\infty$ as if it were a real number.

If you need more convincing about that last sentence, let me ask what you would say is $\infty+1$? Probably you will say it is still $\infty$. But if $$\infty+1=\infty\ ,$$ and if you claim that you can do arithmetic with $\infty$ according to "normal" rules", then we can subtract $\infty$ from both sides and get $1=0$. I hope it's now clear that you can't do this kind of thing.

Final comment: if you know that an integral of this type converges (i.e., has a finite value), then you can calculate it in pretty much the way you suggest. But only if you already know that it converges!

Theorem. If $\displaystyle\int_{-\infty}^\infty f(x)\,dx$ converges, then $$\int_{-\infty}^\infty f(x)\,dx=\lim_{a\to\infty}\int_{-a}^a f(x)\,dx\ .$$