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I have isolated my troubles with solving a larger question to this one part. For $a,b\in\mathbb{R}>1$

What is

$$ \lim_{x\rightarrow 1} \left[\text{Ei}(a\ln(x))-\text{Ei}(b\ln(x))\right] $$

I have no idea where to even begin with this. It arose as I was trying to evaluate the definite integral of

$$ \iint_A x^ydxdy $$

When the x bound was 1, you get the above shown indeterminate form that I can't figure out how to evaluate. Does Ei have some sort of property that allows it to become other functions in some way?

An equivalent form of the limit is

$$ \lim_{x\rightarrow 1} \left[\text{Li}(x^a)-\text{Li}(x^b)\right] $$

ExodusSolis
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Hint: if $x\to 1$ then $a\ln(x)\to 0\;$ so that you are searching $\operatorname{Ei}(a\ln x)-\operatorname{Ei}(b\ln x)$ for parameters going to $0$.
The result should be : $\displaystyle\ln\frac ab$ from the expansion $6.6.1$ at DLMF or here : $$\operatorname{Ei}(z)=\gamma +\frac 12 \left(\ln z-\ln\frac 1z\right) +\sum_{n=1}^{\infty} \frac{z^n}{n!n}$$

Raymond Manzoni
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    I am not sure why your equation is different from the DLMF but using that I can confirm that $\ln(\frac{a}{b})$ is the correct answer as the constant cancels, the infinite sum goes to zero with the input being 0 when x=1, then what is left is the difference of natural logs which can be simplified such that the internal natural logs cancel, resulting in $\ln\frac{a}{b}$. Many thanks! – ExodusSolis Feb 01 '24 at 22:19
  • Glad it helped @ExodusSolis! I indicated Wolfram's formula that doesn't need the $z\ge 0$ hypothesis and handles the complex logarithm (the same at the end). Excellent continuation, – Raymond Manzoni Feb 01 '24 at 23:50