Prove that $\int_{0}^{\fracπ2}(\log(\tan x))^{2n}dx=\left (\frac {π} {2} \right )^{2n+1} \left ( \frac {d^{2n} \sec(z)}{d z^{2n}} \right ) _{z=0}$

Question: Prove that for $$n\in Z^{+}$$ $$\int_{0}^{\fracπ2}(\log (\tan x))^{2n}dx=\left (\frac {π} {2} \right )^{2n+1} \left ( \frac {d^{2n} \sec(z)}{d z^{2n}} \right ) _{z=0}$$

I used fourier expansion of $$\log(\tan x)$$ function $$\log(\tan x)=-2\sum_{k=0}^{\infty}\frac{\cos(2(2k+1)x)}{2k+1}$$ For $$x\in(0,\frac{π}{2})$$

Which makes hard to evaluate summation.Also Integration by parts doesn't work since integrals becomes big to obtain some reduction formual.I could't figure out any other method to proceed further.

• It may be useful to know that $$\left( {\frac{{d^{2n} \sec z}}{{dz^{2n} }}} \right)_{z = 0} = ( - 1)^n E_{2n} .$$ This is coming from the Taylor formula and the know series expansion dlmf.nist.gov/4.19.E5
– Gary
Jun 22 '20 at 19:42
• You can't just take the power of an infinite or even finite sum in this way. You are effectively saying that $(x+y)^n=x^n+y^n$. Jun 22 '20 at 19:44
• Oh,I did that wrong by mistake Jun 22 '20 at 19:47

Famously (using these),$$\int_0^{\pi/2}\tan^{2s-1}xdx=\frac12\operatorname{B}(s,\,1-s)=\frac12\Gamma(s)\Gamma(1-s)=\frac{\pi}{2}\csc(\pi s),$$so$$\int_0^{\pi/2}\tan^{2z/\pi}xdx=\frac{\pi}{2}\csc\left(\frac{\pi(2z/\pi+1)}{2}\right)=\frac{\pi}{2}\sec z.$$Finally, apply $$\left(\frac{\pi}{2}\frac{d}{dz}\right)^{2n}=\left(\frac{\pi}{2}\right)^{2n}\frac{d^{2n}}{dz^{2n}}$$ before setting $$z=0$$.