# Integral $\int_0^y \exp\left( \alpha x + \frac{1}{1-\beta e^{\gamma x}}-\frac{1/\beta }{ 1-\beta e^{-\gamma x}}\right)dx$

I'm studying the integral \begin{align*} \int_0^y \exp\left( \alpha x + \frac{1}{1-\beta e^{\gamma x}}-\frac{1/\beta }{ 1-\beta e^{-\gamma x}}\right)dx \end{align*} with constants $$\alpha,\beta,\gamma$$ (choosen such the fractions are well-defined).

I evaluated the integral numerically but Wolfram Alpha is able to solve special cases, see here or here using the exponential integral, $$\text{Ei}(x)=-\int_{-x}^\infty \frac{e^{-u}}{u}du=\int_{-\infty}^x \frac{e^u}{u}du$$.

Any ideas how the above integral relates to the exponential integral or other special functions?

The special cases Wolfram Alpha solves are $$\int \exp\left( x + \frac{1}{1-2 e^{x}}\right)dx=\frac{1}{2}\left(\text{Ei}\left(\frac{1}{1-2e^x}\right)+e^{1/(1-2e^x)}(2e^x-1)\right)$$ and $$\int \exp\left( x + \frac{1}{1-3 e^{-x}}\right)dx=3e\left(e^{3/(e^x-3)}-\text{Ei}\left(\frac{3}{-3+e^x}\right)+e^{1/(1-3e^{x})}(e^x-6)\right).$$

Here is a free step-by-step solution for the special cases.

The shape of the answer hints you that in the first case Alpha uses the substitution $$u:=\dfrac1{1-2e^x}$$, from which $$2e^x=1-\dfrac1u$$, and

$$\int \exp\left( x + \frac{1}{1-2 e^{x}}\right)dx=\frac12\int\left(\frac1{u-1}-\frac1u\right)e^udu.$$

• I agree with $2e^x=1-\frac{1}{u}$ but isn't $\frac{du}{dx}=u^22e^x=u(u-1)$ and thus, $$\int e^{x+u}dx = \int \frac{1}{2}\left(1-\frac{1}{u}\right)e^u\frac{du}{u(u-1)}=\frac{1}{2}\int \frac{e^u}{u^2}du$$
– Alex
Sep 18 '20 at 13:09
• @Alex: probably. Then you can integrate by parts.
– user65203
Sep 18 '20 at 13:19
• Would you know how to deal with the case if there are three terms in the exponential (first equation in the question) rather than the one special case?
– Alex
Sep 18 '20 at 13:57
• @Alex: if WA can't, I can't, and only Ramanujan can.
– user65203
Sep 18 '20 at 14:33

Assume $$\beta,\gamma\neq0$$ for the key case.

\begin{align}\int_0^ye^{\alpha x+\frac{1}{1-\beta e^{\gamma x}}-\frac{\frac{1}{\beta}}{1-\beta e^{-\gamma x}}}~dx&=\int_1^{e^{\gamma y}}u^\frac{\alpha}{\gamma}e^{\frac{1}{1-\beta u}-\frac{\frac{1}{\beta}}{1-\frac{\beta}{u}}}~d\left(\dfrac{\ln u}{\gamma}\right)\\ &=\dfrac{1}{\gamma}\int_1^{e^{\gamma y}}u^{\frac{\alpha}{\gamma}-1}e^{\frac{1}{1-\beta u}-\frac{u}{\beta(u-\beta)}}~du\\ &=\dfrac{1}{\gamma}\int_1^{e^{\gamma y}}u^{\frac{\alpha}{\gamma}-1}e^\frac{1-u^2}{(\beta u-1)(u-\beta)}~du\\ &=\dfrac{1}{\gamma}\int_1^{e^{\gamma y}}\sum\limits_{n=0}^\infty\dfrac{u^{\frac{\alpha}{\gamma}-1}(1-u)^n(u+1)^n}{n!(\beta u-1)^n(u-\beta)^n}~du\end{align}

The issue is far more complicated, even the integral $$\int_1^{e^{\gamma y}}\dfrac{u^{\frac{\alpha}{\gamma}-1}(1-u)^n(u+1)^n}{(\beta u-1)^n(u-\beta)^n}~du$$ we decomposed should at least relates to Lauricella hypergeometric series.

• I see that you use the substitution $u=e^{\gamma x}$ with $du=\gamma udx$ but how does that new integral relate to the exponential integral, $\text{Ei}(x)$ or perhaps some other special function?
– Alex
Sep 18 '20 at 11:48