We can use the relationship: $$\beta(j,k)=\dfrac{\Gamma(j)\Gamma(k)}{\Gamma(j+k)} ; j,k >0 $$
Setting $j=z$ and $k=1-z$, we obtain : $$ \Gamma(z)\Gamma(1-z) = \beta(z,1-z); 0 < z < 1$$
Now, $\beta(z,1-z)$=$\int_0^1x^{z-1}(1-x)^{-z}dx$ = $2\int_0^{\pi/2}tan^{n}(t)dt$, after the substitution $x=sin^2(t)$ and $n=2z-1$
Next, we define $I_{n}=\int_0^{\pi/2}tan^{n}(t)dt$ ; $-1 < n < 1 $.
$I_{n}=\int_0^{\pi/2}tan^{n}(t)dt=\int_0^{\pi/2}tan^{n-2}(t)(sec^2(t)-1)dt = \int_0^{\pi/2}tan^{n-2}(t)sec^2(t)dt - I_{n-2}$. After the substitution $tan(t)=u$, we get: $$ I_{n} + I_{n-2} = \int_0^\infty u^{n-2}du$$
However, the limits pose a problem, as the integral evaluates $\infty$ [keeping in mind the domain of $n$]: Which is incorrect. We can see this through reverse engineering: $I_{n}$ should be equal to $\dfrac{\pi}{2sin(\pi z)}=\dfrac{\pi}{2sin(\pi (\dfrac{n+1}{2})}=\dfrac{\pi}{2} sec\dfrac{n\pi}{2}$. Therefore,
$I_{n} + I_{n-2} =\dfrac{\pi}{2} sec\dfrac{n\pi}{2} + \dfrac{\pi}{2} sec\dfrac{(n-2)\pi}{2} =0$.
Which leads me to two questions, the major one being in bold:
- What have I done wrong to obtain $\infty$ as the value of the integral instead of $0$?
- Even if I somehow prove $I_{n}+I_{n-2}=0$, I don't see how I can extrapolate the value of $I_{n}$ from this. The only conclusion I can draw from this is that $I_{n}$ might be some function of $sin^{l}(\dfrac{n\pi}{2}+k)$, where $l$ is odd, since increasing n by 2 adds pi to the argument which flips the sign. Which seems to suggest that my approach isn't correct, and so, Is there any other method which allows me to prove that $I_{n}=\dfrac{\pi}{2} sec\dfrac{n\pi}{2}$?