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I have a solved exercise that estimate the divergence of $$\int _0^1\frac{1}{\log\left(2-t\right)}dt$$ using $\frac{1}{\log\left(2-t\right)}\sim \frac{1}{1-t}$ for $t\to 1$. So then $$\int _0^1\frac{1}{1-t}dt=\lim _{x\to 1}\left(\int _0^x\frac{1}{1-t}\right)dt=\lim _{x\to 0}\left(-\log(x)\right)=+\infty $$ So then also the first integral diverges. But where does this come from? I mean, yeah, they are asymptotic for $x\to 1$, but what does have it to do with the integral? Because it start from $0$ up to $1$, and so it is an interval, for which the two functions are clearly not equal. Does this have to do with generalized integrals?

Bernard
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giancarletto
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1 Answers1

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Let $f$ and $g$ be continuous functions on $[a,b)$ and let $f(x)\geq 0$ on that interval. Assume that $\lim_{x\to b^{-}}\frac{f(x)}{g(x)}=c$ exists and $0<c<\infty$. Then the integral $\int_a^b f(x)\,dx$ converges if and only if $\int_a^b g(x)\,dx$ converges.

This is the fact that you're using to conclude that since $\int_0^1 \frac{1}{1-t}\,dt$ diverges and $\frac{1}{1-t}\sim \frac{1}{\log(2-t)}$ as $t\to 1^-$, the given integral has to diverge as well.

There's also a solution with direct comparison using the inequality $\log y\leq y-1$ for all $y>0$: $$\log(2-t)\leq 1-t \Rightarrow \frac{1}{\log(2-t)}\geq\frac{1}{1-t}\,\,\text{for}\,\, t\in[0,1)$$ and the divergence follows.

bjorn93
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