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I want to integrate $$ \int_0^{\infty}dx\,e^{-ax}\frac{1-(2x)^b}{1-2x} $$ where $a,b>0$. Any ideas about how to proceed?

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HINT

Possibly factor into a series: $$ 1-z^b = (1-z)\sum_{k=0}^{b-1} z^k $$

Then your integral becomes $$ \int_0^\infty \mathrm{d}x e^{-ax} \sum_{k=0}^{b-1} (2x)^k $$ which you can compute as the sum of the integrals since the sum is finite...

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    $b$ need not be a natural number. – 2017-02-24
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    @SimplyBeautifulArt i agree but this will give you an approach for natural numbers, and seeing the form of the answer, i would try to speculate that it looks the same for reals -- but that would need proof, of course... – 2017-02-24
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    You could try an infinite series of Gamma functions. – 2017-02-24
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    @gt6989b thanks for the input but what I am interested in is the $b$ non integer case – 2017-02-24
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    @AnarchistBirdsWorshipFungus there does not seem to be a wealth of suggestions for your question yet -- why not try to see what the answer for an integer $b$ is and hope to prove that real $b$ has the same form? – 2017-02-24
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    @gt6989b I have done that, and the answer you get gives you a function that has an imaginary part for some values of $b$, which shouldn't be the case since the result should be real, so you cannot extrapolate from the integer case... – 2017-02-24
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    Using this suggestion I found $$-{a^{b+1}-2^{b+1}\over a-2}$$ – 2017-02-24