# Calculate the integral $\int_0^{\infty} \frac{1}{\lambda + \left(\frac{1}{\lambda} - \lambda\right) e^{-s}} ds$

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

• No further elaboration required.. Thanks! – HolyMonk Dec 12 '17 at 15:36
• It amazes me that Mathematica couldn't figure this out – HolyMonk Dec 12 '17 at 15:38
• @HolyMonk don;t forget to checkmark for future readers – Guy Fsone Dec 12 '17 at 15:38
• I will, but I have to wait 5 minutes, you were too fast! – HolyMonk Dec 12 '17 at 15:39
• 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. – 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

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