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.
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.
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 $x>0$.
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,