While reviewing improper integrals, a thought of the definition of logarithmic function came to me.
$\ln(x) =\int_{1}^{x}\frac{1}{t}dt$
Looking at the definition, I wanted to test for its convergence as the upper limit approached $0$. Having an understanding of the direct and limit comparison tests, I tried comparing the integrand with functions like $\frac{1}{t^2}$, $\frac{1}{\sqrt{t}}$, etc., only to arrive at no conclusion. Is there any way in which I can understand the convergence of $\int_{1}^{0}\frac{1}{t}dt$?
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$\begingroup$Here's a more detailed exposition of what I put in a comment.
First we prove that $\ln (a b) = \ln a + \ln b$. Observe that $$\int_{1}^{ab} \frac{\mathrm{d}t}{t} = \int_{1}^{a} \frac{\mathrm{d}t}{t} + \int_{a}^{ab} \frac{\mathrm{d}t}{t} .$$ Set $u = a/t$. Then \begin{align*} \int_{1}^{a} \frac{\mathrm{d}t}{t} + \int_{a}^{ab} \frac{\mathrm{d}t}{t} & = \int_{1}^{a} \frac{\mathrm{d}t}{t} + \int_{1}^{b} \frac{\mathrm{d}u}{u} \\ & = \ln a + \ln b. \end{align*}
Take any $x \in (0, 1)$. You know by observation that $\ln(x) = - \int_{x}^{1} \frac{\mathrm{d}t}{t} < 0$, so $\lim_{n \to \infty} \log \left( x^n \right) = \lim_{n \to \infty} n \log x = - \infty$. So it definitely can't converge. To show it diverges to $ - \infty$ requires the additional observation that $\ln$ as defined is increasing. So if $a < b$, then $\ln a < \ln b$.
So let $M \in \mathbb{N}$, and let $N$ such that $N \log x < - M$. Then if $0 < y < x^{N}$, then $\ln y < N \ln x < - M$. Thus $\lim_{y \to 0^{+}} \ln y = - \infty$.
$\endgroup$ 1 $\begingroup$You can write this $\ln(x) =\int_{1}^{x}\frac{1}{t}dt$ as $\ln(x) =-\int_{x}^{1}\frac{1}{t}dt$ for $x$ less than $1$ and you are interested in those $x$´s- Since you want to know what happens at $0$ we are interested in $\lim_{\varepsilon \to 0^+} (-\int_{0+ \varepsilon}^{1}\frac{1}{t}dt)$.
We can write $[0,1]$ as $ \bigcup_{n=1}^{\infty} [\dfrac {1}{n}, \dfrac {1}{n+1}] $. On every interval of the form $[\dfrac {1}{n}, \dfrac {1}{n+1}]$ choose the minimal value that $\dfrac {1}{t}$ attains to make elements of the Riemann sum $\dfrac {1}{t_n} \cdot (\dfrac {1}{n} - \dfrac {1}{n+1})$ as small as possible. $\dfrac {1}{t}$ attains minimal value on $[\dfrac {1}{n}, \dfrac {1}{n+1}]$ at the point $\dfrac {1}{n}$ so you have that near zero integral $\int_{x}^{1}\frac{1}{t}dt$ is always greater than $1 \cdot \dfrac {1}{2} + 2 \cdot \dfrac{1}{6}+...+n \cdot \dfrac {1}{n(n+1)}$ for every $n \in \mathbb N$ (for bigger $n$`s just get closer to zero) and this is harmonic series in disguise and because it diverges so your integral diverges to $+\infty$ and because of that you have that $\lim_{\varepsilon \to 0^+} (-\int_{0+ \varepsilon}^{1}\frac{1}{t}dt)$ diverges to $- \infty$.
This can be made more rigorous but you should be able to fill in minor details if needed.
$\endgroup$ $\begingroup$Here's a way to do it: You want to evaluate $$ \lim_{x\to 0^+}\int_1^x\frac{1}{t}dt.$$ Let $t = \frac{1}{u^2}$ so that $dt= -2\frac{1}{u^3}du$ so you get $$ -\lim_{x\to 0^+}2\int_0^{1/\sqrt{x}}\frac{1}{u}du$$ (this in particular shows that $\log(x) = - 2\log(1/\sqrt{x})$ which could also be obtained by the basic log identities).
Now the upper limit is going to infinity and you can use a comparison to the harmonic series to show divergence, provided you've shown that by some more elementary means.
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