I'm not a math major, and I've found myself in need of urgent help dealing with the following integral, pls.
$$ \int_{-2}^1 \frac{1}{x^2} dx $$
It looks simple at first except the anti derivative becomes undefined at $x=0$. So how do you deal with integrals like this? Also is there any book that deals w integrals like this?
On
Integrals like this where the function isn't defined over the entire domain are called improper integrals.
In the case where the point where the function isn't defined is one of the bounds, let's say $b,$ we define $\int_a^b f(x) dx = \lim_{t \to b^-} \int_a^t f(x) dx.$ If this limit exists and is finite, the integral converges to that limit, otherwise the integral diverges and has no real value.
In your case, the singularity occurs inside the interval, so we handle this by splitting the integral up:
$$\int_{-2}^1 \frac{1}{x^2} dx = \int_{-2}^0 \frac{1}{x^2} dx + \int_0^1 \frac{1}{x^2} dx$$
If both integrals converge, the original integral converges to their sum. Otherwise, it diverges.
So, we evaluate the first integral:
$$\int_{-2}^0 \frac{1}{x^2} dx = \lim_{b \to 0^-} \int_{-2}^b \frac{1}{x^2} dx = \lim_{b \to 0^-} \left[-\frac{1}{2} - \frac{1}{b}\right] = \infty$$
Which means that our integral diverges and has no real value.
One way of interpreting $$\int_{-2}^1 \frac{dx}{x^2} $$ is as a sum of two improper integrals: $$\int_{-2}^0 \frac{dx}{x^2} +\int_0^1 \frac{dx}{x^2} \\ = \bigg (-\frac 1x \bigg | _{-2}^0 + \bigg(-\frac 1x \bigg|_0^1 \\ = -\lim_{x\to 0^-} \frac 1x -\frac 12 +\lim_{x\to 0^+ }\frac 1x -1 \\ = 2\lim_{x\to 0^+}\frac 1x -\frac 32 $$ This thing goes to $+\infty$.