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I have some trouble trying to prove that the Cauchy principal value of this important integral $$ \int_0^x \frac{dx}{\ln x} $$ exists if $ x > 1 $. I thought about the expansion $$ \frac{1}{\ln(1+x)} = \frac{1}{x} + \frac{1}{2} - \frac{x}{12} + R(x) $$

where $R(x) = o(x)$ when $x$ approaches zero. Hence for every $ \varepsilon > 0 $ there is $\delta > 0$ such that for every $x \in (-\delta, +\delta)$ we have $|R(x)| < \varepsilon$.

The Cauchy principal value of the integral above may be written as

$$ \lim_{\varepsilon \rightarrow 0 } \left[\int_0^{1-\varepsilon} \frac{dx}{\ln x} + \int_{1+\varepsilon}^x \frac{dx}{\ln x} \right] $$

I thought the expansion above would allow the values just before 1 be eliminated by the values just after 1, but I have no idea how to advance further.

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1 Answer 1

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$$\text{PV}\int_{1-\varepsilon}^{1+\varepsilon}\frac{dx}{\log x}=\text{PV}\int_{0}^{\varepsilon}\frac{1}{\log(1+x)}+\frac{1}{\log(1-x)}\,dx=\int_{0}^{\varepsilon}\frac{\log(1-x^2)}{\log(1+x)\log(1-x)}\,dx $$ and the RHS equals $$ \int_{0}^{\varepsilon} 1+O(x^2) \,dx = \varepsilon + O(\varepsilon^3).$$ In other terms, $\frac{1}{\log(1+x)}$ has a simple pole at the origin with residue $1$, hence we may integrate through such singularity in principal value. This raises an interesting question (maybe): is there a closed form in terms of elementary and standard special functions for $$ \text{PV}\int_{0}^{2}\frac{dx}{\log x}=\int_{0}^{1}\frac{\log(1-x^2)}{\log(1+x)\log(1-x)}\,dx=\int_{0}^{+\infty}\left(\frac{1}{\log(2-e^{-t})}-\frac{1}{t}\right)e^{-t}\,dt \approx 1.04516378\;? $$

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