Question

Let $f\left(n\right)=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}.$ Then show that $f\left(n\right)={\int }_0^{\pi /2}\cot \left(\frac{\theta }{2}\right)(1-{\cos }^n\theta )d\theta$

Answer 1

$\mathrm{Let}g\left(x\right)=1+x+x^2+\cdots +x^{n-1}=\frac{x^n-1}{x-1}$

$\therefore f\left(n\right)={\int }_0^{1}g\left(x\right)dx={\int }_0^1\frac{x^n-1}{x-1}dx$

Put $x=\cos \theta$

$\therefore dx=-\sin \theta d\theta$

$\therefore f\left(n\right)={\int }_{\pi /2}^0\frac{({\cos }^n\theta -1)(-\sin \theta )}{(\cos \theta -1)}d\theta$

$={\int }_0^{\pi /2}\frac{(1-{\cos }^n\theta )2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}d\theta$

$={\int }_0^{\pi /2}\cot \left(\frac{\theta }{2}\right)(1-{\cos }^n\theta )d\theta$