How to solve this integration or does this have a closed form?

Question

The integral I am dealing with is below.

I need to find the closed-form expression of this integral.

$$\int_0^\infty \ln\left(1+\frac{A}{1+B+Cx}\right)\frac{e^{-x/M}}{M}\,dx.$$

Here, $A$, $B$, $C$ and $M$ are constants.

How can I do it?

Answer 1

We will assume $C>0$. You can redo all of the following if it's negative. Replace $1+B+Cx$ with $y$ to get \begin{align*} 1+B+Cx&=y \\ C\,dx &= dy \\ dx &=\frac{dy}{C}\\ -\frac{x}{M}&=-\frac{y-1-B}{CM}=-\frac{y}{CM}+\frac{1+B}{CM} \\ \int_0^\infty \ln\left(1+\frac{A}{1+B+Cx}\right)\frac{e^{-x/M}}{M}\,dx &=\exp\left(\frac{1+B}{CM}\right)\int_{1+B}^\infty \ln\left(\frac{y+A}{y}\right)\,\frac{e^{-y/(CM)}}{CM}\,dy\\ &=\frac{e^{(1+B)/(CM)}}{CM}\int_{1+B}^\infty [\ln(y+A)-\ln(y)]\,e^{-y/(CM)}\,dy. \end{align*} That's about as far as I can go without machine help. The Wolfram Dev Platform will compute this, if the real part of $CM>0.$ The result is quite messy, and involves the Meijer G function. I get this:

$$C M \left(e^{A/(CM)} G_{1,2}^{2,0}\left(\frac{A+B+1}{C M}\Bigg| \begin{array}{c} 1 \\ 0,0 \\ \end{array} \right)-G_{1,2}^{2,0}\left(\frac{B+1}{C M}\Bigg| \begin{array}{c} 1 \\ 0,0 \\ \end{array} \right)-\log (A+B+1) e^{A/(CM)}+\log (A+B+1) e^{-(1+B)/(CM)}+e^{A/(CM)} \log \left(\frac{A+B+1}{C M}\right)-e^{A/(CM)} \log \left(\frac{1}{C M}\right)+\log (B+1) \left(1-e^{-(1+B)/(CM)}\right)-\log \left(\frac{B+1}{C M}\right)+\log \left(\frac{1}{C M}\right)\right). $$