Question

What is `int_(-pi)^pi ((x^2+1)/(x^3cosx)) dx` ?

Answer 1

`int_(-pi)^pi ((x^2+1)/(x^3cosx)) dx = 0` or does it?

Actually, it does not converge. It is of indeterminate form `oo - oo`.

Explanation:

Note that:

• The integration range is symmetrical about `0`.

• The function `x^2+1` is even.

• The function `x^3` is odd.

• The function `cosx` is even.

Hence the function `(x^2+1)/(x^3cos x)` is odd.

So the positive and negative contributions to the integral will cancel out.

Hence:

`int_(-pi)^pi ((x^2+1)/(x^3 cos x)) dx = 0`

graph{(x^2+1)/(x^3cosx)(sqrt(pi^2-x^2)/sqrt(pi^2-x^2)) [-3.2, 3.2, -10, 10]}

`color(red)"EXCEPT"`

The integral does not converge

The problem is that neither of the integrals:

`int_(-pi)^0 (x^2+1)/(x^3cos x) dx" "` `" "int_0^pi (x^2+1)/(x^3cos x) dx`

converges.

As `x->0` the integrand behaves similarly to `1/x^3`, whose integral is `-1/(2x^2)`.

So if `a > 0` is small then:

`int_0^a (x^2+1)/(x^3 cos x) dx ~~ int_0^a 1/x^3 dx`

`color(white)(int_0^a (x^2+1)/(x^3 cos x) dx) = [-1/(2x^2)]_0^a`

`color(white)(int_0^a (x^2+1)/(x^3 cos x) dx) = -1/(2a^2)-(-1/(2(0^+)^2))`

`color(white)(int_0^a (x^2+1)/(x^3 cos x) dx) = +oo`

Hence the integral over the whole `[-pi, pi]` range is effectively `oo-oo`, i.e. indeterminate.

Answer 2

`int_(-pi)^pi (x^2+1)/(x^3cosx)dx = 0` in principal value.

Explanation:

The function:

`f(x) = (x^2+1)/(x^3cosx)`

does not have a primitive in terms of elementary functions. However you do not need to actually solve the integral.

Just note that `f(x)` is an odd function, that is:

`f(-x) = ((-x)^2+1)/((-x)^3cos(-x)) = -(x^2+1)/(x^3cosx) = -f(x)`

In general, the definite integral of an odd function over an interval that is symmetric with respect to the origin must be null, since:

`int_(-a)^a f(x)dx = int_(-a)^0 f(x)dx + int_0^a f(x)dx`

Substitute in the first integral: `t= -x` so that for `x-> -a`, `t->a`

`int_(-a)^a f(x)dx = -int_(a)^0 f(-t)dt + int_0^a f(x)dx`

Inverting the limits of integration of the first integral changes its sign:

`int_(-a)^a f(x)dx = int_(0)^a f(-t)dt + int_0^a f(x)dx`

and as `f(-t) = -f(t)`:

`int_(-a)^a f(x)dx = -int_(0)^a f(t)dt + int_0^a f(x)dx = 0`

Now in this particular case the integral:

`int_0^pi (x^2+1)/(x^3cosx)` is not convergent, but you can evaluate:

`I(epsilon)= int_(-pi)^(-pi/2-epsilon) (x^2+1)/(x^3cosx) dx + int_(-pi/2+epsilon)^(epsilon) (x^2+1)/(x^3cosx) dx int_epsilon^(pi/2-epsilon) (x^2+1)/(x^3cosx)dx +int_(pi/2+epsilon)^pi (x^2+1)/(x^3cosx)dx`

and see that `I(epsilon)` is identically null, so that:

`lim_(epsilon->0) I(epsilon) = 0`

which means that the integral is null in the sense of Cauchy's principal value.