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I want to integrate $$ \int_0^{\infty}dx\,e^{-ax}\frac{1-(2x)^b}{1-2x} $$ where $a,b>0$. My The naive approach was to consider $b$ to be an integer, in which case you get a truncated geometric series above that gives you a sum of gamma functions. Nonetheless, by doing this you get a function that when tried for $b$ non integer gives you a number with an imaginary part, so the naive approach of solving for integer $b$ and hoping that the result can be generalized does not work. Any suggestions?

Any ideas about how to proceed?

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    $\begingroup$ Mathematica says $\frac{1}{2} e^{-\frac{a}{2}-i \pi b} \left(\Gamma (b+1) \Gamma \left(-b,-\frac{a}{2}\right)-e^{i \pi b} \Gamma \left(0,-\frac{a}{2}\right)\right)$, which is a hint. Note that Mathematica gets clumsy integrating around branch cuts, and sometimes ends up with a nonzero imaginary part erroneously. $\endgroup$ Feb 24, 2017 at 18:41
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    $\begingroup$ This integral diverges because of a pole at $x=1/2$. If you want a "principal value integral" you should say so. $\endgroup$ Feb 24, 2017 at 18:50
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    $\begingroup$ @GeraldEdgar no pole, it is an avoidable singularity $\endgroup$ Feb 24, 2017 at 18:54
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    $\begingroup$ You are right, no ingularity. I had $1-2x^b$ in error. $\endgroup$ Feb 24, 2017 at 22:55

2 Answers 2

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The Mathematica 11 code

i = Integrate[Exp[-a*x]*(1 - (2*x)^b)/(1 - 2*x), {x, 0, Infinity}, 
  PrincipalValue -> True, Assumptions -> a > 0 && b > 0]

outputs $$\frac{1}{2} e^{-\frac{a}{2}} \left((-1)^b \Gamma (b+1) \Gamma \left(-b,-\frac{a}{2}\right)+\text{Ei}\left(\frac{a}{2}\right)+i \pi \right) $$ Let us check it:

N[i /. {a -> 2, b -> 1}]

$ 0.5\, +0. i $

Addition.

N[i /. {a -> 2.1, b -> 3/2}]

$0.68219\, +7.770178924974121\,\,10^{-17} i $

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  • $\begingroup$ Try it for b non integer and you will see an imaginary part $\endgroup$ Feb 24, 2017 at 19:27
  • $\begingroup$ You are not right. See the addition in my answer. The very small imaginary part appears because of errors when calculating. $\endgroup$
    – user64494
    Feb 24, 2017 at 19:39
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Using a certain exponential integral from Maple, it comes out as $$ {\frac {-b!\,{2}^{b+1}+ \left( -{a}^{b+1}{\rm Ei}_1 \left(-a/2 \right) + \left( b+1 \right) !\,{2}^{b+1}{\rm Ei}_{b+2} \left(-a/2 \right) \right) {{\rm e}^{-a/2}}}{2{a}^{b+1}}} $$ It works for fractional $b$, interpreting $b! = \Gamma(b+1)$ as usual.

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