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So I was evaluating this improper integral, and found the antiderivative to be $e^{1/x}(1-\frac{1}{x})$. How would I evaluate it from $0$ to $-1$? In other words, what would $\frac{1}{0}$ be?
$$\int_{-1}^0 {\frac{e^{1/x}}{x^3}}dx$$

I know there is a similar post to this question: Improper Integral of $\int_{-1}^0 \frac{e^\frac{1}{x}}{x^3}dx$

I've read it, but it doesn't really help me...

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liya77
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1 Answers1

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Notice that the $0$ in “ $\dfrac10$ ” is slightly negative, so we are actually dealing with

$\displaystyle\lim_{x\to0^-}e^{1/x}\bigg(1-\frac1x\bigg)$, where the first part goes towards $e^{-\infty}=0$, and the latter

towards $\infty$. Applying l'Hopital, the two quantities balance each other out, and

the limit is finite.

Lucian
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  • @ Lucian How do you know it's slightly negative and not positive? Or is there two situations? – liya77 Mar 10 '15 at 02:30
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    @liya77: Because you are integrating on $(-1,0)$, as opposed to, say, $(0,1)$, so the upper integration limit tends to $0$ from below or from the left. – Lucian Mar 10 '15 at 02:35