How do you evaluate the integral #int tanxdx# from 0 to #pi# if it converges?

2 Answers
Dec 3, 2016

# int_0^pi tanx dx # is convergent and # int_0^pi tanx dx = 0#

Explanation:

#tan x# has a discontinuity when #x=pi/2# so for a thorough robust solution we would need to split the integral as follows;

#int_0^pi tanx dx = lim_(delta x rarr (pi/2)^(+)) int_0^(delta x) tanx dx + lim_(delta y rarr (pi/2)^(-)) int_(delta y)^pi tanx dx#

However as #tanx# is an odd function with period #pi# observer that it has rotation symmetry about #x=pi/2#, and so even though #tan x# is discontinous at #x=pi/2# we have:

# int_0^pi tanx dx = 0# and is therefore convergent

Dec 3, 2016

The integral does not converge.

Explanation:

In order to evaluate #int_0^pi tanx dx#, we must evaluate both of

#int_0^(pi/2) tanx dx# #" "# and #" "# #int_(pi/2)^pi tanx dx#

and add the results.

#int_0^(pi/2) tanx dx = lim_(brarr pi/2) int_0^(pi/2) tanx dx# #" "# if the limit exists (is finite)

# = lim_(brarrpi/2) {: -lncosx]_0^b# (integrate by substitution)

# = lim_(brarrpi/2) (-lncosb + lncos0)#

# = lim_(brarrpi/2) (-lncosb + ln1)#

# = lim_(brarrpi/2) -lncosb#

# = oo#.

The limit does not exist.

So, #int_0^(pi/2) tanx dx# diverges.

Therefore #int_0^pi tanx dx# diverges.

Bonus material

There is a similar notion, called the Cauchy principal value, that has us, in a sense, evaluate the two improper integrals at once.

For #tanx# on #[0,pi]# it is defined by

#lim_(epsilonrarr0^+) (int_0^(pi/2+epsi) tanx dx + int_(pi/2+epsi)^pi tanx dx)#.

This limit is #0#.

Second example:

#int_-oo^oo x dx# does not converge because neither #int_-oo^c x dx# nor #int_c^oo x dx# converges

The Cauchy principal value is again #0# because for this integral the value is

#lim_(ararroo) int_-a^a x dx = lim_(ararroo) 0 = 0#