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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.