\( \int _ { - 1 } ^ { 1 } \frac { e ^ { 1 / x } } { x ^ { 3 } } d x \)

To evaluate the integral \( \int_{-1}^{1} \frac{e^{1/x}}{x^3} \, dx \), we first recognize that the integrand, \( \frac{e^{1/x}}{x^3} \), has issues at \( x=0 \) where it tends to infinity (the function is not defined there). Because of this singularity, we must consider the nature of the integral and whether it converges. To handle the singularity, we can split the integral: \[ \int_{-1}^{1} \frac{e^{1/x}}{x^3} \, dx = \int_{-1}^{0} \frac{e^{1/x}}{x^3} \, dx + \int_{0}^{1} \frac{e^{1/x}}{x^3} \, dx \] Near \( x = 0 \), the behavior of the integrand differs on either side. For \( x \to 0^+ \), \( e^{1/x} \) grows to \( +\infty \) rapidly, while for \( x \to 0^- \), \( e^{1/x} \) approaches zero since it will tend towards 0 exponentially fast. As such, the first integral from -1 to 0 converges, while the second from 0 to 1 diverges. In summary, the integral does not converge since one part diverges while the other does not. Thus, the overall evaluation reveals the integral to diverge.