The improper integral $\displaystyle \int_{-\infty}^\infty x \, dx$ is

A. divergent by comparison to $\displaystyle \int_{-\infty}^\infty x e^{-x} \, dx$.
B. divergent by comparison to $\displaystyle \int_{-\infty}^\infty \sqrt{x} \, dx$.
C. convergent since the area to the left of $x = 0$ cancels with the area to the right of $x = 0$.
D. convergent since it equals $\displaystyle \lim_{t \to \infty} \int_{-t}^t x \, dx = \lim_{t \to \infty} \left( \frac{t^2}{2} - \frac{(-t)^2}{2} \right) = 0$.
E. convergent since it equals $\displaystyle \lim_{a \to -\infty} \int_a^0 x \, dx + \lim_{b \to \infty} \int_0^b x \, dx = -\infty + \infty = 0$.
F. divergent since $\displaystyle \int_{-\infty}^0 x \, dx$ is convergent and $\displaystyle \int_0^{\infty} x \, dx$ is divergent.
G. divergent since both integrals $\displaystyle \int_{-\infty}^0 x \, dx = -\infty$ and $\displaystyle \int_0^{\infty} x \, dx = +\infty$ are divergent.