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I am working trying to integrate this function...

$$\int_0^{\infty } \frac{e^{-x}}{2 \left(1-\lambda \left(1-e^{-x} (x+1)\right)\right)} \, dx$$

but I can't find a way to do it. However, a similar one does...

$$\int_0^{\infty } \frac{x e^{-x}}{2 \left(1-\lambda \left(1-e^{-x} (x+1)\right)\right)} \, dx = \frac{-\log (1-\lambda)}{2 \lambda }$$

Is there any way to use the second result to integrate the first?

PiE
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  • How did you arrive at the complex value of the second integral? I don't see a reason why $i$ should appear in there ... – MrYouMath Mar 01 '17 at 13:47
  • I used Maple to compute that – PiE Mar 01 '17 at 13:51
  • That doesn't make sense, the integrand is real and you are integrating over a real interval, why should there be something complex in the result? Or is your $\lambda \in \mathbb{C}$? – MrYouMath Mar 01 '17 at 13:52
  • Ok - you are correct - you don't need that value - I removed the complex component – PiE Mar 01 '17 at 14:21
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    @PMF I am almost certain that the answer is $\frac{1}{2-2\lambda}$. But I haven't found a way to prove it yet. – GaussTheBauss Mar 01 '17 at 15:27
  • @GaussTheBauss I tried numerically evaluation the integral using your equation and the answer seems off a bit...With $\lambda$ = 0.5, numerically, your answer gives 1.0 - numerically, I get 0.712686 – PiE Mar 01 '17 at 16:13
  • @PMF Maple gives 1.0 exactly for $\lambda = .5$. – GaussTheBauss Mar 01 '17 at 16:14
  • @ PMF g := proc (y) options operator, arrow; int(exp(-x)/(2-2y(1-exp(-x)(x+1))), x = 0 .. infinity) end proc;

    g(1/2); 1

     – GaussTheBauss Mar 01 '17 at 16:16
  • Hmm..I plugged that formula into my system and I got 0.712686...Not sure what is going on – PiE Mar 01 '17 at 16:43
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    Note that $$(e^{-x} (1 + x))' = -xe^{-x}$$ – Zaid Alyafeai Mar 01 '17 at 16:54
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    This shows the second integral is logarithmic. – Zaid Alyafeai Mar 01 '17 at 16:58
  • This may sound "crazy"...But compared to the 1st integral, it seems like the 2nd integral takes an "expectation" of the first. Maybe there is way using something like "iterated expectations" (or something like that) to recover the first integral from the second... – PiE Mar 01 '17 at 17:09
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    might be helpful: http://math.stackexchange.com/questions/45745/interesting-integral-related-to-the-omega-constant-lambert-w-function/207639#207639 – tired Mar 01 '17 at 18:18
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    @GaussTheBauss this is what WA finds for $\lambda=1/2$ http://www.wolframalpha.com/input/?i=NIntegrate%5BExp%5B-x%5D%2F(1%2BExp%5B-x%5D(x%2B1)),%7Bx,0,10000%7D%5D – tired Mar 01 '17 at 18:27
  • I get that value as well: 0.587137 for $\lambda = 0.5$ – PiE Mar 01 '17 at 18:54
  • @tired - thanks for the reference - I'll check that out... – PiE Mar 01 '17 at 18:56

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