I have the following equation

$$\int_0^1 \frac{a^x -1}{a-1} dx = r,$$

where the integral evaluates as:

$$\frac{1}{1-a} + \frac{1}{\log(a)} = r.$$

I would like to solve for a but this is proving to be quite complex. Any Suggestions?


1 Answer 1


There is no closed solution for $a$ in terms of "elementary" functions, but there is a solution with (generalized) Lambert W functions as follows:

Let $a = e^{v}$ then $\frac{1}{1-e^{v}} + \frac{1}{v} = r$ or $e^{-v} = \frac{r}{r-1}\frac{v- 1/r}{ v- 1/(r-1)}$ where this is of the general form

$e^{-c v} = a_0 \frac{v- q}{ v- s}$ with real constants $c, a_0, q,s$. This is a generalized form of the standard Lambert W function, a reference in wikipedia is here. So you can take it from there to assess solutions of this special form. The article "Maignan, Aude; Scott, T. C. (2016). "Fleshing out the Generalized Lambert W Function". SIGSAM. 50 (2): 45–60" cited in Wikipedia can be found here.


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