QuestionFind the integral: $$ \int_0^{\infty} \frac{1}{\lambda + \left(\frac{1}{\lambda} - \lambda\right) e^{-s}} ds, $$ here $\lambda \in ]0,1[$ is an arbitrary number.

Sidenote In my eyes this integral seems to be fairly easy to solve, but I can't find it (and mathematica also can't solve it).

I do know that integrals can be solved using Complex Analysis but my memory of this approach is very vague so I can't decide whether this is applicable to this problem.


Hint:$$ \frac{1}{c+be^{-s}} = \frac{e^s}{ce^s+b}=\color{red}{\left(\frac{1}{c}\log(ce^s+b)\right)'=}.$$

Then with $c=\lambda$ and $b=\lambda-\frac{1}{\lambda}$ find $$\int_0^{\infty} \frac{1}{\lambda + \left(\frac{1}{\lambda} - \lambda\right) e^{-s}} ds,$$ Which obviously does not converges

  • $\begingroup$ No further elaboration required.. Thanks! $\endgroup$ – HolyMonk Dec 12 '17 at 15:36
  • $\begingroup$ It amazes me that Mathematica couldn't figure this out $\endgroup$ – HolyMonk Dec 12 '17 at 15:38
  • $\begingroup$ @HolyMonk don;t forget to checkmark for future readers $\endgroup$ – Guy Fsone Dec 12 '17 at 15:38
  • $\begingroup$ I will, but I have to wait 5 minutes, you were too fast! $\endgroup$ – HolyMonk Dec 12 '17 at 15:39
  • $\begingroup$ I found it odd that the integral doesn't converge but there was a typo in my question, the $e^{-s}$ was supposed to be $e^s$, but this integral is solved in exactly the same manner as the way you solved this integral. $\endgroup$ – HolyMonk Dec 12 '17 at 15:54

Please note that integral doesn't converge. My approach without considering the interval of lamdda $]0,1[$

Answer is doesn't converge and indefinite integral gives $\log((e^s -1)x^2 +1)/x $ as the answer.

Proceed yourself thereafter

  • $\begingroup$ He is not looking at the convergence. by the way you are right it does converge +1:) $\endgroup$ – Guy Fsone Dec 12 '17 at 15:39
  • $\begingroup$ @Guy Fsone Thanks $\endgroup$ – Chen Guo Dec 12 '17 at 15:40
  • $\begingroup$ @ChenGuo: Thanks for your answer but the other one is even better than yours ;)! $\endgroup$ – HolyMonk Dec 12 '17 at 15:47
  • $\begingroup$ @HolyMonk okay. $\endgroup$ – Chen Guo Dec 12 '17 at 17:08

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