<s>[INST] How to calculate this exponential integral? 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!
 [/INST] 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. 
</s>