How to calculate this exponential integral?

Question

I am trying to calculate

$\int_0^{+\infty}\log_2(1+ax)\frac{1}{a}e^{-\frac{x}{a}}dx$

by using the integration by parts method like:

\begin{equation} \begin{aligned} \int_0^{+\infty}\log_2(1+ax)\frac{1}{a}e^{-\frac{x}{a}}dx &= -\int_0^{+\infty}\log_2(1+ax)(e^{-\frac{x}{a}})'dx\\ &=\log_2(1+ax)e^{-\frac{x}{a}}|_0^{+\infty} + \int_0^{+\infty}\frac{a}{\ln2(1+ax)}e^{-\frac{x}{a}}dx\\ &= \int_0^{+\infty}\frac{a}{\ln2(1+ax)}e^{-\frac{x}{a}}dx = \frac{a}{\ln2}\int_0^{+\infty}\frac{1}{(1+ax)}e^{-\frac{x}{a}}dx\\ &= \frac{a}{\ln2}\int_0^{+\infty} \frac{-\frac{1}{a^2}}{-\frac{1}{a^2}(1 + ax)}e^{-\frac{x}{a}}dx = \frac{-\frac{1}{a}}{\ln2}\int_0^{+\infty} \frac{1}{-\frac{1}{a^2} - \frac{x}{a}}e^{-\frac{x}{a}}dx\\ &= \frac{-\frac{1}{a}}{\ln2}\int_0^{+\infty}\frac{e^{-\frac{x}{a}}e^{-\frac{1}{a^2}}}{(-\frac{1}{a^2} - \frac{x}{a})e^{-\frac{1}{a^2}}}dx \\ &= \frac{-\frac{1}{a}e^{\frac{1}{a^2}}}{\ln2}\int_0^{+\infty}\frac{e^{-\frac{x}{a}-\frac{1}{a^2}}}{(-\frac{x}{a}-\frac{1}{a^2})}dx\\ \end{aligned} \end{equation}

How can I further integrate the $\int_0^{+\infty}\frac{e^{-\frac{x}{a}-\frac{1}{a^2}}}{(-\frac{x}{a}-\frac{1}{a^2})}dx$ in order to get a closed form expression?

Thank you in advance!

Answer 1

You are almost done. Continuing from where you left: $$I =\int_{0}^{\infty} \frac{e^{-x/a-1/a^2}}{-x/a-1/a^2}\, dx$$ Substituting $u = \frac{x} {a} + \frac1{a^2} \implies du = \frac1{a} \, dx$ gives us: $$I = a \int_{\frac1{a^2}}^{\infty} - \frac{e^{-u}}{u} \, du$$ which is the exponential integral $E_1(u)$.

Hope you can take it from here.