I have a fairly good understanding of improper integrals, but this question has been bugging me for a while since I got it wrong on an exam. So here's the integral:
$$\int_{-1}^0 \frac{(e^{1/x})}{x^2}\ dx.$$
So the first thing to do is to recognise that it is improper. That must be because $e^{1/0}$ is not defined, correct?
So proceeding in a fashion like any other improper integral problem. Change the limit $0$ to a $b$ and taking the limit as $b$ approaches zero, like this:
$$\lim\limits_{b \to 0} \int_{-1}^b \frac{(e^{1/x})}{x^2}\ dx.$$
So now that it is proper, we can evaluate the integral, using a $u$-substitution of $u=1/x$. As $-du=1/{x^2} \, dx$ we find $$-\lim_{b \to 0} \int_{-1}^b e^{u}\ du.$$ which the integral will equal the same as the derivative
Plugging $u$ back in gives $$-\lim_{b \to 0} e^{{1/x}}\ \Big|_{-1}^{b}.$$ thus $$-\lim_{b \to 0} e^{{1/b}} + \lim_{b \to 0} e^{{1/(-1)}}. $$
But since you plug zero in for $b$ you get a division by zero error. Can anyone help?
You seem to understand roughly what is going behind the integral, but let me nitpick for a moment. First, bounds should change when you make the substitution:
$$\int_{-1}^{b} \frac{e^{1/x}}{x^2} \, dx = -\int_{-1}^{1/b} e^u \, du = \int_{1/b}^{-1} e^u \, du = e^{-1} - e^{1/b}.$$
A superficial explanation is as follows: The new variable in general does not range over the same interval as the original variable does. So the bounds should also be replaced accordingly to match this new range.
Now we want to take limit as $b \to 0^-$. I guess it is not hard to be convinced from the graph $y= 1/b$
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that $1/b \to -\infty$ as $b \to 0^-$. So we have
$$ \lim_{b\to0^-} \int_{-1}^{b} \frac{e^{1/x}}{x^2} \, dx = \lim_{b\to0^-} \big( e^{-1} - e^{1/b} \big) = e^{-1} - 0 = \frac{1}{e}. $$