Improper Integral of $\int_{-1}^0 \frac{e^\frac{1}{x}}{x^3}dx$

Question

I have a question for The Improper Integral of $\int_{-1}^0 \frac{e^\frac{1}{x}}{x^3}dx$ That's what i have done $u=\frac1x$ $du=\frac{-1}{x^2}$ After integrated by parts I had $e^{\frac1x}(1-\frac1x)$

So the $\lim_{t\rightarrow 0^-} [e^\frac{1}{t}(1-\frac1t) -e^{-1}(1+1)]$

How can I find $\frac{e^\frac1t}{t}$?

Please help

You probably mean you want to find the $\lim_{t\to 0-} e^{1/t}/t$ right? — Umberto, Dec 30 '13 at 13:49
This could help http://math.stackexchange.com/questions/55247/the-rapidity-of-the-exponential-function-towards-infinity — Umberto, Dec 30 '13 at 13:55
You can consider the following hint: try this substitution $s=-1/t$. Then Your limit becomes: $\lim_{s\to +\infty} e^{-s}s$... Now should be easier... — Umberto, Dec 30 '13 at 14:00

score 1 · Accepted Answer · answered Dec 30 '13 at 20:34

1

$$\int_{-1}^{0^{^-}}\frac{e^\frac1x}{x^3}dx=-\int_{-1}^{-\infty}e^tt^3\frac{dt}{t^2}=\int_{-\infty}^{-1}te^tdt=-\int_\infty^1(-u)e^{-u}du=-\int_1^\infty ue^{-u}du=$$

$$=\left[\frac{u+1}{e^u}\right]_1^\infty=0-\frac2e=-\frac2e$$

answered Dec 30 '13 at 20:34

Lucian

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Improper Integral of $\int_{-1}^0 \frac{e^\frac{1}{x}}{x^3}dx$

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