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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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This integral is just $\int_0^1 x^{-(a+ibA)} \, dx$. – Patrick Da Silva Aug 6 '11 at 16:19
What is the motivation for this problem? Is it homework, or is it something you actually came across in some real-world application? Why do you have both a constant $b$ and a constant $A$, when only their product appears in the integral? Please show some partial work so we can see that you've made some effort. I'm downvoting because this fits exactly within the criterion shown in the pop-up help box for the downvote button: "This question does not show any research effort..." – Ben Crowell Aug 6 '11 at 17:01
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 $x$. – Christian Blatter Aug 6 '11 at 17:38
up vote 3 down vote accepted

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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Why can't you comment? Do you see the "add comment" button? – Patrick Da Silva Aug 6 '11 at 16:26
@Patrick Da Silva : I can only comment on my own answer as I am doing now. Other than that, "add comment" is disabled for me. Besides, it didn't feel like an answer but anyway, not a big deal. I see that your answer was faster than mine so sorry for not noticing on time. – user13838 Aug 6 '11 at 16:39
@Patrick: You need 50 reputation to comment (see – joriki Aug 6 '11 at 16:40
Don't worry for putting a second answer, it's not a big deal at all. It happens all the time. – Patrick Da Silva Aug 6 '11 at 17:19
@joriki : XD I guess I didn't stay there long enough to notice... lol – Patrick Da Silva Aug 6 '11 at 17:19

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,

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