Now, I already know the complement formula and I also know a proof that can be presented using the Euler product forms of the Gamma function. What I am curious about is, can one obtain the formula using the integral forms of the Gamma and Beta functions?(adsbygoogle = window.adsbygoogle || []).push({});

I will show my work so far. I started with the identity that implies that the Beta function is an analytic continuation of the binomial coefficients, that is,

[tex]B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}[/tex]

Next, I plugged in 1-a for b to obtain

[tex]B(a,1-a)=\Gamma(a)\Gamma(1-a)[/tex]

Note that the right-hand side is exactly what the complement formula tells the value of. So, it remains to obtain a simple expression for the Beta function on the left. This is easily seen to be equal to

[tex]\Gamma(a)\Gamma(1-a) = 2\int_{0}^{\pi/2}\tan^{2a-1}(\theta)d\theta[/tex]

So my question is, can we evaluate this integral? (that is, without using the product representations.)

I already know the value of the integral from the complement formula, and I verified that the path I trace is correct in the sense that Wolfram gives the exact complement formula when evaluating this integral. I want to know if it is possible to do it by hand.

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# The complement formula of the Gamma function

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