Answer
$$
\int_{0}^{1} \frac{e^{1 / x}}{x^{3}} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{e^{1 / x}}{x^{3}} $ has the vertical asymptote $x =0 $. The infinite discontinuity occurs at the left endpoint of $ [0, 1] $.
The given improper integral
$$
\int_{-1}^{0} \frac{e^{1 / x}}{x^{3}} d x= \infty
$$
is divergent
Work Step by Step
$$
\int_{0}^{1} \frac{e^{1 / x}}{x^{3}} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{e^{1 / x}}{x^{3}} $ has the vertical asymptote $x =0 $. Since the infinite discontinuity occurs at the left endpoint of $ [0, 1] $ , we must use part (b) of Definition 3 :
$$
\begin{split}
\int_{0}^{1} \frac{e^{1 / x}}{x^{3}} d x & = \lim _{t \rightarrow 0 ^{+}} \int_{t}^{1} \frac{e^{1 / x}}{x^{3}} d x \\
& =\lim _{t \rightarrow 0^{+}} \int_{t}^{1} \frac{1}{x} e^{1 / x} \cdot \frac{1}{x^{2}} d x \\
& =\lim _{t \rightarrow 0^{+}} \int_{1 / t}^{1 } u e^{u}(-d u) \quad\left[\begin{array}{c}{u=1 / x} \\ {d u=-d x / x^{2}}\end{array}\right] \\
& =-\lim _{t \rightarrow 0^{+}} \int_{1 / t}^{1} u e^{u}(d u),
\quad\quad \text{integration by parts } \\
& =-\lim _{t \rightarrow 0^{+}}\left[(u-1) e^{u}\right]_{1 / t}^{1} \\
& = \lim _{t \rightarrow 0^{+}}\left[0 +\left(\frac{1}{t}-1\right) e^{1 / t}\right] \\
& =\lim _{s \rightarrow \infty}(s-1) e^{s} , \quad\quad[s=1 / t] \\
& = \infty
\end{split}
$$
Thus the given improper integral
$$
\int_{0}^{1} \frac{e^{1 / x}}{x^{3}} d x = \infty
$$
is divergent.
