Question

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

Answer 1

`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

Answer 2

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`