0
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ 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 $\endgroup$
    – Gary
    Jun 22 '20 at 19:42
  • $\begingroup$ You can't just take the power of an infinite or even finite sum in this way. You are effectively saying that $(x+y)^n=x^n+y^n$. $\endgroup$ Jun 22 '20 at 19:44
  • $\begingroup$ Oh,I did that wrong by mistake $\endgroup$
    – Paras
    Jun 22 '20 at 19:47
4
$\begingroup$

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 $z=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.