One integral involving integrals exponential and logarithmic function Is there a closed-form solution for the integral 
$$ \int_{0}^{\infty}\log_{2}(1+ax)\cdot e^{-bx} \; \mathrm dx $$ 
with $a, b \geq 0$?
If there is no closed-form solution, whether there is an approximation for the integral?
Thanks.
 A: You may express your integral in terms of a known special function, here in terms of the incomplete gamma function defined by

$$
\Gamma(\alpha,z):=\int_\alpha^\infty u^{z-1}e^{-u}du
$$ 

many of its properties are recorded here.
Indeed, you have
$$
\begin{align}
 \int_{0}^{\infty}\log_{2}(1+ax)\cdot e^{-bx} \; \mathrm dx &= \frac{e^{b/a}}{a\ln2}\int_1^{\infty}\ln v\cdot e^{-bu/a} \; \mathrm dv \quad (v = 1+ax)\\\\
&=\left[  \frac{e^{b/a}}{a\ln2}\ln v\left(-\frac abe^{-bv/a}\right)\right]_1^{\infty}+\frac{e^{b/a}}{b\ln2}\int_1^{\infty}v^{-1}\cdot e^{-bv/a} \; \mathrm dv\\\\
&=\frac{e^{b/a}}{b\ln2}\int_{b/a}^{\infty}u^{0-1}\cdot e^{-u} \; \mathrm du\qquad (u=bv/a)\\\\
&=\frac{e^{b/a}}{b\ln2}\:\Gamma\left(0,\frac ba\right).
\end{align}
$$
A: Ignoring limits, you can probably simplify this enough to see whether there's something obvious. First, $\log_2$ is a constant multiple of $\ln$, so I'm going to ignore the $ln_2$. Let's see where that goes:
\begin{align}
\int ln(1 + ax) e^{-bx}~ dx 
&= a\int \ln(1 + y) e^{-\frac{b}{a}y}~ dy & \text{ substitute $y = ax$}\\  
&= a\int \ln(z) e^{-\frac{b}{a}(z-1)}~ dz & \text{ substitute $z = 1+y$}\\  
&= a\int \ln(z) e^{-\frac{b}{a}z} e^{-\frac{b}{a}}~ dz \\  
&= a e^{-\frac{b}{a}}\int \ln(z) e^{-\frac{b}{a}z}~ dz \\  
\end{align}
so now we're down to something a little less messy:
$$
\int \ln(z) e^{-cz}~ dz
$$
I ought to know whether that's integrable in elementary terms, but I don't. The obvious choice here -- integration by parts with $f = \ln z$ and $g = e^{-cz}$ -- doesn't seem to lead anywhere nice. But it's possible that since you want a definite integral, this can actually be computed by some residue trick. I defer to people who do complex analysis daily to see whether there's something obvious here. 
