# How can I solve this integration or does it have a closed form solution?

The integral I am dealing with is below.

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

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

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

How can I do it in Mathematica?

Mathematica can do it if you add assumption:

\$Version
(*"12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)"*)

Integrate[Log[1 + a/(1 + b + c*x)]*Exp[-x/m]/m, {x, 0, Infinity},Assumptions -> {a > 0, b > 0, c > 0, m > 0}]

(*E^((1 + b)/(c m)) ExpIntegralEi[-((1 + b)/(c m))] -
E^((1 + a + b)/(c m)) ExpIntegralEi[-((1 + a + b)/(c m))] +
Log[(1 + a + b)/(1 + b)]*)


In LaTeX code:

$$\int_0^{\infty } \frac{\ln \left(1+\frac{a}{1+b+c x}\right) \exp \left(-\frac{x}{m}\right)}{m} \, dx=e^{\frac{1+b}{c m}} \text{Ei}\left(-\frac{1+b}{c m}\right)-e^{\frac{1+a+b}{c m}} \text{Ei}\left(-\frac{1+a+b}{c m}\right)+\ln \left(\frac{1+a+b}{1+b}\right)$$

for:$$\{a>0,b>0,c>0,m>0\}$$

• Here, E is Exp? – dipak narayanan Apr 26 '19 at 15:59
• @dipaknarayanan .Yes. – Mariusz Iwaniuk Apr 26 '19 at 16:03