Question

How do you determine if the improper integral converges or diverges $\int{\left( \dfrac{1}{3x}-6 \right)dx}$ from negative infinity to 0?

Answer 1

To solve the given question, we should know the integration of some of the functions that are given below: we should know that the integration of $\dfrac{1}{ax}$ is $\dfrac{\ln \left( ax \right)}{a}$, and integration of a constant function k is $kx$. We should also know that, to evaluate a two-sided improper integral, split it into two integrals and express each as a limit. For the given question we will break integral as from negative infinity to -1 and from -1 to 0.

Complete step by step solution:
We are asked to determine if the improper integral converges or diverges $\int{\left( \dfrac{1}{3x}-6 \right)dx}$ from negative infinity to 0. As we already know that, to evaluate a two-sided improper integral, split it into two integrals and express each as a limit.
We have $\int\limits_{-\infty }^{0}{\left( \dfrac{1}{3x}-6 \right)dx}$, splitting the improper limit as from negative infinity to -1 and from -1 to 0. We can also write it as $\displaystyle \lim_{a \to -\infty }\int\limits_{a}^{-1}{\left( \dfrac{1}{3x}-6 \right)dx}+\displaystyle \lim_{b\to 0}\int\limits_{-1}^{b}{\left( \dfrac{1}{3x}-6 \right)dx}$.
As $\dfrac{1}{3x}$ is similar to $\dfrac{1}{ax}$, its integration will be $\dfrac{\ln \left( 3x \right)}{3}$. And the integration of -6 will be $-6x$.
Using these integrations in the above limits, we get

& \Rightarrow \displaystyle \lim_{a \to -\infty }\left( \dfrac{\ln \left| 3x \right|}{3}-6x \right)_{a}^{-1} \\\ & \Rightarrow \displaystyle \lim_{a \to -\infty }\left[ \left( \dfrac{\ln 3}{3}+6 \right)-\left( \dfrac{\ln 3a}{3}-6a \right) \right] \\\ & \Rightarrow \left[ \left( \dfrac{\ln 3}{3}+6 \right)-\left( \infty +\infty \right) \right]\to -\infty \\\ \end{aligned}$$ As this integral is diverging, the whole function will also diverge. **Note:** Here, we stopped by evaluating only one integral because it was diverging, we know that one integral diverges, the whole expression will diverge. We can also evaluate limits for the other integral, but the result will be the same. Hence, the given integral will diverge from negative infinity to 0.