Q:

Help on an integral.

I have to finish following integral
$$\int_0^{+\infty} \log_2(1+a x) (1-e^{-\frac{x}{2}})^{b-1} \frac{1}{2}e^{-\frac{x}{2}} dx$$
After a week of work on this, I found it may be impossible to have a integral analytically. So I think I need to have an approximation of this.
The shape of the integrated function is as following. With fixed b and varying a:

With fixed a and varying b:
Could you give me a hint on the approximation of the integral?

A:

You can approximate the integral using Gauss-Laguerre quadrature (http://en.wikipedia.org/wiki/Gauss%E2%80%93Laguerre_quadrature) for integrals of the form
$$
\int_0^{\infty} f(x)e^{-x}dx \approx \sum_{i=1}^n w_i f(x_i),  
$$
where $x_i,i=1,...,n$ is the root of the $n$-th degree Laguerre polinomial $L_n(x)$ and $
w_i,i=1,...,n$ is
$$
w_i=\frac{x_i}{(n+1)^2(L_{n+1}(x_i))^2}.
$$
Picking $n=1$ results in an approximation which, albeit crude, has the advantage of being a simple (and interpretable) function of $a$ and $b$, as $L_1(x)=-x+1,L_2(x)=1/2(x^2-4x+2),x_1=1$ and $w_1=1$. Defining $f(u)=\log_2(1+2au)(1-e^{-u})^{b-1}$, the integral can be approximated as follows:
$$
\int_0^{\infty} f(u)e^{-u}du\approx 1\cdot f(1)=\log_2(1+2a)(1-e^{-1})^{(b-1)}\quad (*)
$$
This table shows a few instances of the correct integral $I(a,b)$ and approximation $\hat I(a,b)$.
a b I I^
0 -1 0 0 
0 0 0 0 
0 1 0 0 
1 0 2.650694 2.507374 
1 1 1.331479 1.584963 
1 2 0.9013049 1.001887 
2 1 1.934489 2.321928 
2 2 1.268749 1.467738 
2 3 0.9494982 0.9277877 
3 2 1.508186 1.774587 
3 3 1.117553 1.121753 
3 4 0.8908656 0.709083 

More accurate numerical results can be obtained with $n=2$ or larger but the simplicity of $(*)$ will be lost. Hope this helps!

