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I am stuck on how to integrate the following: $$\int_0^1 x^{a-1} (1-x)^{-a} dx$$ where $a \in (0,1)$. I am aware that this is a variant of the Euler gamma/beta functions and will be equal to $\pi/\sin(a \pi)$. However, I would like to derive this result without using this. After doing a substitution with $e^t = 1/(1-x)$, the integral transforms into $$\int_0^\infty \frac{dt}{(e^t-1)^{1-a}}.$$ I tried using a standard keyhole integral with the branch cut on the positive real line, but I run into issues with the fact that the residues apparently aren't defined. How do I tackle integrals like this?

On a similar note, what is the general technique for integrating something more general, like $$\int_0^1 x^a (1-x)^b dx$$ for real valued $a,b$? I tried doing something with a modified keyhole around the singularities at $0$ and $1$ which agrees with this reference (https://math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf), but I keep getting answers that are nonsensical.

Editing to add: the integral in question was originally posed on my department's complex analysis qualifying exam from a couple years ago, with the specific instructions to not use the gamma and beta functions and I was not successful with this.

homie o'morphic
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    for the second one thats more general you can just use beta function https://math.stackexchange.com/questions/188190/how-to-integrate-int-01xa1-xbdx but if you want to do it the complex analysis way then fair enough too lol – Max0815 Aug 02 '23 at 22:30
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    I don't think there is a pure complex analytic way to do the general case of $a,b$ since the answer involves the Gamma function and you will somehow have to manipulate your contour integral into an equivalent form of the Gamma function integral - you may as well then just use the standard evaluation of the Beta function with the Gamma function which has an easy real analytic proof – FShrike Aug 02 '23 at 22:33
  • That makes sense; I saw the first link, but it was not helpful for me... For reference, the integral in question was originally posed on my department's complex analysis qualifying exam from a couple years ago, with the specific instructions to not use the gamma and beta functions. – homie o'morphic Aug 03 '23 at 02:49

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$\newcommand{\d}{\,\mathrm{d}}$As commented, I don't think there's any morally interesting way to approach the general case with complex analysis. I feel sure this is a duplicate but I couldn't find any - for using a purely complex analytic technique - so here goes.

For the specific case where the answer is of the form $\pi\csc\pi a$, you just use a dogbone contour! I have explained this technique many times on this site before, here are three recent examples - the first is most relevant to you.

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FShrike
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    These are all very helpful, thank you!! I realize now that the "dogbone contour" was what I should have been trying to search for instead. I don't know how I did not stumble across these while trying to search for this integral on both stackexchange and on Google, these are quite similar. I will keep this question up anyway in case another misinformed person ends up searching the same wrong things in trying to figure out how to solve this problem. – homie o'morphic Aug 03 '23 at 02:47