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Let $\delta$ be an element of $(0,1)$. I need to find an upper bound for the integral from $0$ to delta of $-1/\log$. Clearly, this is an improper integral, so I let a be an element of $(0,\delta)$ and try to then find the limit as a goes to $0$ from the right of this integral. $-1/\log$ has no elementary primitive, so I need to use a different method. So far I have shown that the function $-1/\log$ is increasing on $(0,1)$ and hence that the integral from a to delta of $-1/\log$ is less than or equal to $-(\delta - a) / \log(\delta)$. Thus, as long as I knew that the limit as a goes to $0$ from the right of the integral from a to delta of $-1/\log$ EXISTS, it would follow that this limit would have to be less than or equal to $-(\delta - a) / \log(\delta)$ - I would have found an upper bound for the integral from 0 to delta of $-1/\log$. But I don't know how to show that this limit exists. Help please?

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Consider the function $f:[0,1) \to \mathbb{R}$ defined by $f(0)=0$ and $f(t)=-(\log t)^{-1}$ for $0<t<1$. Then $f$ is obviously continuous, and so for every $b<1$ the function $$ F:[0,b] \to \mathbb{R}, F(x)=\int_x^bf(t)dt $$ is well defined. Furthermore, $F$ is differentiable, and in particular $F$ is continuous. Thus $$ F(0)=\lim_{a\to 0}F(a)=\lim_{a\to 0}\int_a^b-(\log(t))^{-1}dt. $$

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The integral: $$\int{\frac{-1}{\log{x}}}dx = -Li(x) + C$$ Where $Li(x)$ is the Logarithmic Integral.

To prove that the integral exists on the region $[0,\delta]$ where $\delta < 1$, all you have to do is show that the function is continuous and bounded in that range (which it is). Since: $$\lim_{x\rightarrow 0} -1/\log(x) = 0$$

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