When I calculate the integration $\int {{e^{\frac{{ - \lambda x}}{{A - Bx}}}}} dx$ where $A,B,\lambda > 0$ some how Mathematica return the $Ei(x)$ but do not show the step. Could anyone plase explain to me how to arrive at this result ?
Thank you for your enthusiasm !
Asked
Active
Viewed 49 times
0
1 Answers
1
I suppose that this start with $$-\frac{\lambda x}{A-B x}=t \implies x=\frac{A t}{B t-\lambda }\implies dx=-\frac{A \lambda }{(B t-\lambda )^2}$$ to make $$I=-A \lambda\int \frac{ e^t}{(B t-\lambda)^2}\,dt$$ Then $$t=\frac{\lambda -u}{B}\implies I=\frac{A \lambda e^{\frac{\lambda }{B}}}{B}\int \frac{e^{-\frac{u}{B}}}{u^2}\,du$$ Now $u=B v$ $$I=\frac{A \lambda e^{\frac{\lambda }{B}}}{B^2}\int \frac{e^{-v}}{ v^2}\,dv$$ Now, one integration by parts and you are done since $$\int \frac{e^{-v}}{ v^2}\,dv=-\frac{e^{-v}}{v}-\text{Ei}(-v)$$
Claude Leibovici
- 260,315
-
Oh thank you very much ! how could you come up with the ideas of $t = \frac{{\lambda - u}}{B}$ ? and $u = Bv$ – Tuong Nguyen Minh Apr 14 '20 at 09:26
-
@TuongNguyenMinh. The first, just to simplify the denominator.The second to have the simplest argument in the exponential. And third, 64 years of experience !! Cheers :-) – Claude Leibovici Apr 14 '20 at 09:38
-
Oh now I understand your insight ! – Tuong Nguyen Minh Apr 14 '20 at 13:46