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

Question

Paisley Johnson · Accepted Answer

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

Ezra Adams · Answer

The integral does not converge.

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#

Samantha Adkison · Answer

undefined To evaluate the integral $\int_0^{\pi} \tan(x) \, dx$, we first note that $\tan(x)$ is not defined at $x = \frac{\pi}{2}$ where it has an infinite discontinuity. However, within the given bounds $(0, \pi)$, the integral converges.

Using the substitution $u = \cos(x)$, $du = -\sin(x) \, dx$, we have:

$$\int \tan(x) \, dx = \int \frac{\sin(x)}{\cos(x)} \, dx$$

$$= \int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos(x)| + C$$

Evaluating the integral from $0$ to $\pi$:

$$\int_0^{\pi} \tan(x) \, dx = -\ln|\cos(\pi)| + \ln|\cos(0)|$$

$$\ln|\cos(0)| = \ln|1| = 0$$

$$\ln|\cos(\pi)| = \ln|-1| = \ln(1) = 0$$

So,

$$\int_0^{\pi} \tan(x) \, dx = -\ln(1) + \ln(1) = 0$$