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I changed my earlier problem. $\int_{0}^{1}\frac{1}{x^a}\frac{1}{e^{ibA\ln{x}}}dx$

$i$ is imaginary unit, $a$,$b$,$A$ is constant.

Any answer will be appreciated, thank you.

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    I would suggest the substitution $x:=e^{-u}$ \ (\infty> u\geq 0). In this way you don't have to deal with strange things like $x^{a+i b A}$ for variable $x$.2011-08-06

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I still can't comment but isn't this simply equal to

$ \begin{align} \int_{0}^{1}\frac{1}{x^a}\frac{1}{e^{ibA\ln{x}}}dx &= \int_{0}^{1}\frac{1}{x^a}\frac{1}{(e^{\ln{x}})^{ibA}}dx\\ &=\int_{0}^{1}\frac{1}{x^a}\frac{1}{x^{ibA}}dx\\ &= \int_{0}^{1}\frac{1}{x^{a+ibA}}dx\\ \end{align} $ for $x>0$.

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    @joriki : XD I guess I didn't stay there long enough to notice... lol2011-08-06
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This integral is just $\int_0^1 x^{-(a+ibA)} \, dx$. Since you can re-write $ e^{ibA \ln(x)} = (e^{\ln(x)})^{ibA} = x^{ibA}, $ you get the above integral and the result follows quite easily since we know antiderivatives for powers of $x$.

Hope that helps,