Integrating $f(x) = 1/x$ from $x=a$ to $x=\infty$ Can the integration of $f(x)=1/x$ from $x=a > 0 $ to $x=\infty$ ever be finite?
That is, can $\int_{x=a}^{\infty} 1/x$ be finite?
 A: Since $\frac1x \geq 0$ for $x \geq 0$, we have
$$
\int_{x=a}^\infty \frac1x dx \geq \int_{x=a}^N \frac1x dx = \log(N) - \log(a)
$$
for any $N \geq a$, the first integral possibly taking the value $\infty$. Now suppose $\int_{x=a}^\infty \frac1x dx$ is finite and take $N$ with $\log(N) > \log(a) + \int_{x=a}^\infty \frac1x dx$ to derive a contradiction.
A: Let $a>0$. By the change of variables $t=x/a$, we see
$$
\int_{x=a}^\infty \frac1x\,dx=\int_{t=1}^\infty\frac1t\,dt
$$
So the integral is finite for one $a$ if and only if it's finite for every $a$.
A: When integrated, the answer becomes $[\ln(\infty) - \ln(a)]$. 
Assume for argument that $\ln(\infty) = \infty$. Subtracting any finite number from infinity will not leave a finite number. 
You may think your only option left is to make constant 'a' equal to $\infty$. Then you are subtracting infinity from infinity and get .... voila .... lots of arguments. Is the answer zero (a finite number) or some other version of infinity? 
The answer goes back to the "arguable" statement used to start the logic chain, is $\ln(\infty) = \infty$ or is $\ln(\infty)$ undefined. Perhaps that can be your next question.   
