Can someone help me to find a solution for the following integral?

$ \int_{0}^{\infty} \frac{(ax+c)}{(ax+b) ^2} e^{-\frac{x}{b}} dx$

Thank you.

  • $\begingroup$ $$\frac{e^{1/a} \text{Ei}\left(-\frac{1}{a}\right) (a b+b-c)+a (b-c)}{a^2 b}$$ $\endgroup$ – Raffaele Nov 30 '20 at 15:29
  • $\begingroup$ @Raffaele. "Dont give a child a fish but show him how to fish” ( Mao Zedong) $\endgroup$ – Claude Leibovici Dec 1 '20 at 4:36
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    $\begingroup$ @ClaudeLeibovici Et si le garçon voulait plutôt un espadon? $\endgroup$ – Raffaele Dec 1 '20 at 8:55
  • $\begingroup$ @Raffaele. This is a good answer ! Cheers :-) $\endgroup$ – Claude Leibovici Dec 1 '20 at 9:05

Making the problem more general, consider $$I=\int \frac{ (\alpha x+\beta)}{(\gamma x+\delta )^2}\,e^{- \epsilon x}\,dx$$

Concerning the antiderivative first, let $\gamma x+\delta=t$ to make $$I=\frac{e^{\frac{\delta \epsilon }{\gamma }} }{\gamma^2} \int\frac{\alpha t +(\beta \gamma-\alpha \delta)}{t^2} {e^{-\frac{t \epsilon }{\gamma }} }\,dt$$

Now, $t=\frac{\gamma }{\epsilon }y$ to make $$I=\frac{e^{\frac{\delta \epsilon }{\gamma }} }{\gamma^3} \int \frac {\alpha \gamma y +\epsilon(\beta \gamma -\alpha \delta ) } {y^2 }\,e^{-y}\,dy$$

Now, two integrals $$I_1=\int \frac {e^{-y}} y \,dy=\text{Ei}(-y)$$ $$I_2=\int \frac {e^{-y}} {y^2} \,dy=-\text{Ei}(-y)-\frac{e^{-y}}{y}$$

Back to $x$ $$I=\frac{ (\alpha \delta -\beta \gamma )}{\gamma ^2 ( \gamma x+\delta)}e^{- \epsilon x }+\frac{\alpha \gamma +\epsilon(\alpha \delta -\beta \gamma ) }{\gamma ^3 }e^{\frac{\delta \epsilon }{\gamma }} \text{Ei}\left(-\frac{\epsilon }{\gamma }(\gamma x+\delta )\right)$$ Now, for the definite integral, there will be conditions to be fulfilled.

For you specific case, the result would be
$$\int_{0}^{\infty} \frac{(ax+c)}{(ax+b) ^2} e^{-\frac{x}{b}}\, dx=\frac{a (c-b)-e^{\frac{1}{a}} \text{Ei}\left(-\frac{1}{a}\right) (a b+b-c)}{a^2 b}$$ if $$(\Re(a) \Im(b)\neq \Im(a) \Re(b)\lor \Re(a)\geq 0)\land \left(\Re\left(\frac{b}{a}\right)\geq 0\lor \frac{b}{a}\notin \mathbb{R}\right)\land \Re(b)>0$$

  • $\begingroup$ Thank you @ClaudeLeibovici for the detailed explanation! $\endgroup$ – Mina Kay Nak Dec 1 '20 at 13:25
  • $\begingroup$ @MinaKayNak. You are welcome ! It is a nice problem. $\endgroup$ – Claude Leibovici Dec 1 '20 at 13:26

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