integral of power of $\tan(x)$ from $0$ to $\pi/2$

Question

Prove that $$\int_{0}^{\frac{\pi}{2}} \tan^k(x)dx=\frac{\pi}{2}\sec{\frac{\pi k}{2}}$$ for $k < 1$ . I saw this problem for the case where $k=\frac{1}{2020}$ here. Then I checked a few rational values for $k$ like $\frac{1}{2}, \frac{1}{3}$ and they all seemed to satisfy the above equation. Now for proving it, I tried to use the well known fact that $\int_{0}^{\frac{\pi}{2}}\sin^{2p-1}(x)\cos^{2q-1}(x)dx=\frac{\Gamma(p)\Gamma(q)}{2\Gamma(p+q)}$. Substituting $p=\frac{k+1}{2}, q=\frac{1-k}{2}$, we have that $$\int_{0}^{\frac{\pi}{2}} \tan^{k}(x)dx=\frac{\Gamma(\frac{1+k}{2})\Gamma(\frac{1-k}{2})}{2}$$ but I have no idea to compute $\Gamma(\frac{1+k}{2})\Gamma(\frac{1-k}{2})$. Also I considered differentiating both sides of the relation to be proven with respect to $k$, but that leaves to prove that $\int_{0}^{\frac{\pi}{2}}\tan^k(x)\ln(\tan(x))dx=\frac{\pi^2}{4}\sec\frac{\pi k}{2}\tan{\frac{\pi k}{2}}$. Integration by parts may be tried here but that leaves us with expressions which are difficult to handle. Now how do I proceed?

Answer 1

Using $$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$$ from https://en.wikipedia.org/wiki/Gamma_function, one has $$ \Gamma(\frac{1+k}{2})\Gamma(\frac{1-k}{2})=\frac{\pi}{\sin(\frac{1+k}{2}\pi)}= \frac{\pi}{\sin(\frac{\pi}{2}+\frac{k}{2}\pi)}=\pi\sec(\frac{k}{2}\pi)$$