Question

Integral 1/x from 0 to infinity:

\[\int\limits_{0}^\infty\frac{1}{x}dx\]

Answer 1

\[\int\limits_{0}^\infty\frac{1}{x}dx=\lim_{t \rightarrow 0^+}\int\limits_{t}^\infty\frac{1}{x}dx=\lim_{a \rightarrow \infty}\lim_{t \rightarrow 0^+}\int\limits_{t}^a\frac{1}{x}dx\]
\(\ln|x|_t^a=\ln|a|-\ln|t|=\infty-\infty\) = undefined?

Answer 2

(Also, does the order of the limits matter in general? I haven't really done "double limits" before)

Answer 3

1/x is not integrable on that interval because the function approaches 0 as x approaches in infinity and function approaches infinity as x approaches 0 from the positive side. This results in an infinitely increasing area under the curve which means that the function is not integrable.

Of course you could still approximate the area with Riemann Sums because the infinitely increasing area is gonna make very little difference no matter how far we go in the interval (0, infinity). Thus, we can still approximate the area but the function itself is not integrable.

Answer 4

Ok , but can't some functions still converge as the integral approaches infinity, like e^(-x) from x=0 to infinity gives 1, as an example we did. Is this not how we treat improper integrals? Can we actually "eyeball" the function like that o_O? I dunno I'm not sure how you can determine the above by not calculating it. These improper integrals don't seem obvious to me

Answer 5

you can eyeball it because the anti derivative is the log, which goes to infinity as x does, and also goes to minus infinity as x goes to zero

Answer 6

This particular integral will diverge, yes. I'm going to rewrite some of your work, but only out of personal preference:
\[\begin{align*}\int_0^\infty \frac{1}{x}dx&=\int_0^c \frac{1}{x}dx+\int_c^\infty \frac{1}{x}dx&\small c\text{ is some constant, }0<c<\infty\\
&=\lim_{a\to0^+}\int_a^c \frac{1}{x}dx+\lim_{b\to\infty}\int_c^b \frac{1}{x}dx\\
&=\lim_{a\to0^+}\left[\ln|x|\right]_a^c+\lim_{b\to\infty}\left[\ln|x|\right]_c^b\\
&=\ln(c)-\lim_{a\to0^+}\ln|a|+\lim_{b\to\infty}\ln|b|-\ln(c)\\
&=\lim_{b\to\infty}\ln|b|-\lim_{a\to0^+}\ln|a|\\
&=\infty-(-\infty)\\
&=\infty\end{align*}\]

Answer 7

oh i see thank you so much