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How to show that $\operatorname{li}(x)=\operatorname{Ei}(\log x)$?

Here, $\operatorname{li}(x)$ and $\operatorname{Ei}(x)$ are defined as: $$\operatorname{li}(x)=\int_0^x\frac{\mathrm dt}{\log t}$$ $$\operatorname{Ei}(x)=\int_{-\infty}^x\frac{e^t}t\,\mathrm dt$$ I can't find any proof of this expression but I did find many articles stating it, could anyone show me how it could be proven?

ultralegend5385
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Jesús
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1 Answers1

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If you want to prove: $$ \int _{0}^{x} \frac{1}{\log(t)} \, dt==\int _{-\infty}^{\log(x)} \frac{e^{ t }}{t} \, dt $$ Let left-hand side $t=e^{ u }$ So, $$ \int _{0}^{x} \frac{1}{\log(t)} \, dt = \int _{-\infty}^{\log(x)} \frac{1}{u} \cdot e^{ u } \, du $$ Done.

Xu Chang
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