Final answer:
The integral \( \int_1^{\infty} \, de \) is an improper integral that diverges because as the upper limit of integration approaches infinity, the value of the integral grows without bound.
Step-by-step explanation:
The integral in question is \( \int_1^{\infty} \, de \). This is an improper integral because it involves the integration over an infinite interval, from 1 to infinity. To evaluate whether this integral converges or diverges, we must look at its behaviour as it approaches infinity.
We can write the integral as \( \lim_{b \to \infty} \int_1^{b} \, de \). Calculating this integral gives us \( e \Big|_1^b = e^b - e^1 \). As \( b \) approaches infinity, \( e^b \) grows without bound, which means this improper integral diverges.
Therefore, the correct answer is (a) Diverge.