Why does the integral $\int_{-\infty}^ \infty \frac{1}{x^2}dx$ diverge?

Question

The integral $\int_{-\infty}^ \infty \frac{1}{x^2}dx$ according to many websites diverges. But, by manual calculation, the answer is clearly $0$. Why is this the case?

Answer 1

I just realized it's because the value at x = 0 does not exist. It approaches infinity from both sides at x = 0.

Answer 2

Because when $x$ approaches zero (from either side), $y$ approaches infinity in a way that is not converging. You can look at how $1/x^2$ approaches the positive $y$-axis the same way $1/x$ approaches the positive $x$-axis, both are diverging!

Answer 3

One must be extremely careful and must pay close attention while applying the Fundamental Theorem Of Calculus (Note that the function must be continuous if you want to apply this theorem directly) . This is a common example where many (including myself) made mistakes.

A better way to write this integral would be : (To avoid confusion) $$\lim_{a\to 0^-}\int_{-\infty}^a\frac{1}{x^2}\mathrm{d}x+\lim_{b\to 0^+}\int_{b}^{\infty}\frac{1}{x^2}\mathrm{d}x$$ Upon solving, we get: $$\lim_{a\to0^-}-\frac{1}{x}\bigg|_{-\infty}^a+\lim_{b\to 0^+}-\frac{1}{x}\bigg|_{b}^{\infty}$$ $$=\lim_{a\to 0^-}-\left(\frac1a-\frac{1}{-\infty}\right)+\lim_{b\to0^+}-\left(\frac1{\infty}-\frac1b\right)$$ $$=-\lim_{a\to 0^-}\frac1a+\lim_{b\to0^+}\frac1b$$ Which is clearly divergent.