Solving $\int_{-2}^{a} x^{-2} dx$ as a tends to ∞? A-level FP1 I'm currently going through what I don't know from my A-level course (studying with AQA). In FP1 Calculus one of the final textbook questions is as follows:
'A student evaluates $\int_{-2}^a x^{-2} dx$ as $-\frac{1}{a}-\frac{1}{2} $and concludes that $\int_{-2}^∞ x^{-2} dx$ is equal to $-\frac{1}{2}$. Explain why she is incorrect?
Here is my work so far:

My conclusion was that the student's error was describing $-\frac{1}{2}$ as equal to the integral, because the correct description would be that the integral tends to the value $-\frac{1}{2}$. Would I be correct in giving this as an answer? I was wondering if I had possibly made a mistake in the calculations.
The answer in the textbook is: '$\frac{1}{x^2}$ is not defined when x=0 which is part of the interval of integration.' What I gather from this is that 0 is within the limits, and $\frac{1}{0}$ doesn't have a value. However, I am not sure how this would apply to my answer or how I needed to have factored it into my solution.
The way I worked it out was that the integral converges, however, online 'solver' sites say it diverges. Please could somebody clear this up for me by explaining the correct way to go about answering the question.
Thanks in advance,
Nathan
PS: I don't know how to add an image inline (or if this is possible) so apologies for that.
 A: Your statement is not quite accurate, I'm afraid. When we write $\int_a^\infty f(t) \ dt$, this is just shorthand for $\lim_{x \to \infty} \int_a^x f(t) \ dt$. In your example, if we suppose, as you argue, that the integral does tend to $- \frac 12$, since $ \int_a^{\infty} f(t) \ dt$ is defined as what the integral tends to, we could still technically say that the value is $-\frac 12$. This is assuming your reasoning is correct, which it is not.
In reality, the integral does not exist. The reason your reasoning fails is that you ignore the asymptote at $x=0$. You can't integrate "over" the asymptote. What you should do is split the integral up on both sides of the asymptote and see if both integrals are independently well defined. In this case, they are not. 
In some cases, however, functions with vertical asymptotes are "integrable". ("Integrable" is in quotes because the Riemann integral assumes $f$ to be bounded on $[a,b]$; this is a so-called "improper" integral, which is basically just a limiting procedure)
Example: Suppose $f(x) = \frac{1}{\sqrt{|x|}}$ in $[-1, 1]$ and $f(x) = 0$ for $x \geq 1$. Then $\int_{-1}^{\infty} f(t) \ dt = 4$. Try working this out yourself. 
tl;dr: If a continuous function has a vertical asymptote, you need to split it up and integrate over both sides of the asymptote independently. If the integral is well-defined, you just add up the result. 
A: we have the integral $$\int_{-2}^ax^{-2}dx$$ and we get three cases: the result is 
$$-\frac{1}{2}\frac{a+2}{2a}$$ if $$a<0$$ and $$\infty$$ if $$a=0$$ and $$\infty$$ if $$a>0$$
