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

Answer 1

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

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