How do you evaluate #int cot(x)# from 0 to 2pi?

2 Answers
Sep 5, 2016

The integral diverges.

Explanation:

Consider the function: #cot(x)# has vertical asymptotes at #x=0,pi,2pi...# Thus the integral will diverge.

Sep 5, 2016

This improper integral does not converge.

Explanation:

#cot x# is not defined for #x = 0#, #pi# and #2pi#.

To attempt to evaluate the integral, we need to find the two improper integrals

#int_0^pi cot x dx + int_pi^(2pi) cotx dx#

#int_0^pi cot x dx# is undefined at both limits of integration, so we need to break it somewhere. I'll use #pi/2#

#int_0^pi cot x dx = int_0^(pi/2) cot x dx+int_(pi/2)^pi cot x dx#

#int_0^(pi/2) cot x dx = lim_(ararr0^+) int_a^(pi/2) cotx dx#

# = lim_(ararr0^+) ln(sinx) |_a^(pi/2)#

# = ln(sin(pi/2)) - lim_(ararr0^+) (ln(sina))#

# = ln(1) - lim_(ararr0^+) (ln(sina))#

# = - lim_(ararr0^+) (ln(sina))#

But this limit does not exist, so the integral diverges.

Because one of the integrals involved in #int_0^(2pi) cot x dx# diverges, the integral diverges.