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I need to evaluate this integral as a follow up to a previous question, $\int\limits_{x = D}^{x = \frac{A}{B}} {{e^{\frac{{ - \lambda x}}{{A - Bx}}}}{e^{ - Cx}}dx} $ where $A,B,C,D,\lambda > 0$ are all positive number that is greater than zero

Why the Ei(x) pop up in Integration of ${e^{\frac{{ - \lambda x}}{{A - Bx}}}}$ by Mathematica?

Thank you very much for your good will and enthusiasm !

Attempted effort do far has been unsuccessful because integration by parts make the problem look even more horrible

$\begin{array}{l} u = {e^{ - Cx}} \Rightarrow u' = - C{e^{ - Cx}}\\ v' = {e^{\frac{{ - \lambda x}}{{A - Bx}}}} \Rightarrow v = \frac{{ - A}}{B}{e^{\frac{{ - \lambda x}}{{A - Bx}}}} + x{e^{\frac{\lambda }{B} + \frac{{A\lambda }}{{B\left( {Bx - A} \right)}}}} - \frac{{A{e^{\frac{\lambda }{B}}}\lambda Ei\left( {\frac{{A\lambda }}{{B\left( {Bx - A} \right)}}} \right)}}{{{B^2}}}\\ \left. {uv} \right|_D^{\frac{A}{B}} - \int_{x = D}^{x = \frac{A}{B}} {\left[ { - C{e^{ - Cx}}} \right]} \left[ {\frac{{ - A}}{B}{e^{\frac{{ - \lambda x}}{{A - Bx}}}} + x{e^{\frac{\lambda }{B} + \frac{{A\lambda }}{{B\left( {Bx - A} \right)}}}} - \frac{{A{e^{\frac{\lambda }{B}}}\lambda Ei\left( {\frac{{A\lambda }}{{B\left( {Bx - A} \right)}}} \right)}}{{{B^2}}}} \right]dx \end{array}$

Tuong Nguyen Minh
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  • What is the question here? What have you tried? What problem are you facing? – an4s Apr 14 '20 at 15:09
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    I don't believe there is a closed form without using some function defined as an integral, like $\mathrm{erf}$ or $\mathrm{Ei}$. – aschepler Apr 14 '20 at 15:19
  • oh I mean a closeform in the form of non elementary function is also acceptable – Tuong Nguyen Minh Apr 14 '20 at 22:59
  • Not sure if this helps, but the substitution $u = A-Bx$ yields $$\int_{D}^{A/B} \exp\left(-\frac{\lambda x}{A-Bx} - Cx\right) , dx = \frac{1}{B}\int_0^{A-BD} \exp\left(- \frac{(Cu+\lambda)(A-u)}{Bu}\right) , du$$ – angryavian Apr 14 '20 at 23:37
  • Is there any way to express this in term of special function because when I expand your substitution it boils down to integrate(u-1/u) du on some interval ? – Tuong Nguyen Minh Apr 15 '20 at 00:02

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