Type I (Infinite Intervals)

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When we defined definite integral `int_a^bf(x)dx` we assumed that interval [a,b] is is finite and that function f doesn't have infinite discontinuities. Let's extend integral to the case where interval [a,b] is infinite.

Consider the infinite region S that lies below the curve `y=f(x)` , above the x-axis and to the right of the line x=a.

improper integral of first type

You might think that, since S is infinite in extent, its area must be infinite, but let’s take a closer look. The area of the part of S that lies to the

left of the line x=t is `A(t)=int_a^tf(x)dx=F(t)-F(a)` where `F` is antiderivative of f.

For example, if `f(x)=1/x^2` and `a=1` then `A(t)=int_1^t 1/x^2dx=1-1/t` .

Observe that in this case `lim_(t->oo)A(t)=lim_(t->oo)(1-1/t)=0` .

So, we define the integral of f (not necessarily a positive function) over an infinite interval as the limit of integrals over finite intervals.

Definition of an Improper Integral of Type 1

  1. If `int_a^tf(x)dx` exists for every number `t>=a`, then `int_a^(oo)f(x)dx=lim_(t->oo)int_a^t f(x)dx` provided this limit exists (as a finite number).
  2. If `int_t^b f(x)dx` exists for every number `t<=b`, then `int_(-oo)^(b)f(x)dx=lim_(t->oo)int_t^b f(x)dx` provided this limit exists (as a finite number).
  3. The improper integrals `int_a^(oo)f(x)dx` and `int_(-oo)^bf(x)dx` are called convergent are called convergent if the corresponding limit exists and divergent if the limit doesn't exist.
  4. If both `int_a^(oo)f(x)dx` and `int_(-oo)^a f(x)dx` are convergent then `int_(-oo)^(oo)f(x)dx=int_(-oo)^a f(x)dx+int_a^(oo)f(x)dx` . Note, that any real number can be used. Once again only when BOTH integrals are convergent, convergent is also an integral `int_(-oo)^(oo)f(x)dx` . Otherwise, it is divergent.

Any of the improper integrals in Definition can be interpreted as an area provided that f is a positive function.

Example 1. For which p is integral `int_1^(oo) 1/x^p dx` convergent?

Let's handle two cases: p=1 and `p!=1` .

If p=1 then `int_1^(oo)1/xdx=lim_(t->oo)int_1^t 1/x dx=lim_(t->oo)(ln|x||_1^t)=lim_(t->oo)(ln(t)-ln(1))=oo` .

Thus, when p=1 integral is divergent.

If `p!=1` then `int_1^(oo)1/x^p dx=lim_(t->oo) int_1^t 1/x^p dx=lim_(t->oo)(1/(-p+1)x^(-p+1)|_1^t)=1/(1-p)lim_(t->oo)(1/t^(p-1)-1)` .

If `p>1` then `p-1>0` and `1/t^(p-1)->0` as `t->oo` . Therefore, `int_1^oo 1/x^p dx=1/(p-1)` when p>1.

If `p<1` then `p-1<0` and `1/t^(p-1)->oo` as `t->oo` . Therefore, integral is divergent.

Therefore, `int_1^(oo) 1/x^p dx` is convergent for p>1.

Note the interesting fact: although the curves `y=1/x^2` and `y=1/x` look very similar for x>0, the region under `y=1/x^2` to the right of x=1 has finite area whereas the corresponding region under `y=1/x` has infinite area. Note that both `1/x^2` and `1/x` approach 0 as `x->oo` but `1/x^2` approaches faster than `1/x` . The values of `1/x` don't decrease fast enough for its integral to have a finite value.

Example 2. Evaluate `int_(-oo)^(-1) e^(2x)dx` .

By definition `int_(-oo)^(-1)e^(2x)dx=lim_(t->-oo)int_t^(-1)e^(2x)dx=lim_(t->-oo)(1/2e^(2x)|_t^(-1))=lim_(t->-oo)(1/2e^(-2)-1/2e^(2t))=`

`=1/2e^(-2)-0=1/(2e^2)` .

Example 3. Calculate if possible `int_(-oo)^(oo)1/(r^2+4)dr` .

It is convenient to take `a=0` in definition.

`int_(-oo)^(oo)1/(r^2+4)dr=int_(-oo)^0 1/(r^2+4)dr+int_0^(oo) 1/(r^2+4)dr` .

Now `int_(-oo)^0 1/(r^2+4)dr=lim_(t->-oo)int_t^0 1/(r^2+4)dr=lim_(t->-oo)(1/2tan^(-1)(r/2)|_t^0)=lim_(t->-oo)(1/2tan^(-1)(0/2)-1/2tan^(-1)(t/2))=`

`=lim_(t->-oo)(0-1/2tan^(-1)(t/2))=0-(1/2*(-pi/2))=pi/4`.

Similarly `int_0^(oo) 1/(r^2+4)dr=lim_(t->oo)int_0^t 1/(r^2+4)dr=lim_(t->oo)(1/2tan^(-1)(r/2)|_0^t)=lim_(t->-oo)(1/2tan^(-1)(t/2)-1/2tan^(-1)(0/2))=`

`=lim_(t->oo)(1/2tan^(-1)(t/2)-0)=1/2*pi/2-0=pi/4`.

Since both of these integrals are convergent, the given integral is convergent and `int_(-oo)^(oo)1/(r^2+4)dr=pi/4+pi/4=pi/2` .

Since `1/(x^2+4)>0` , the given improper integral can be interpreted as the area of the infinite region that lies under the curve `y=1/(x^2+4)` and above the x-axis.

Example 4. Find `int_(-oo)^(oo)1/xdx` .

Here we will choose a=1.

`int_(-oo)^(oo)1/x dx=int_(-oo)^1 1/xdx+int_1^(oo) 1/x dx` .

It was shown in example 1 that `int_1^(oo) 1/x dx` is divergent, so we don't need to calculate second integral. Since at least one integral is divergent then `int_(oo)^(-oo) 1/x dx` is also divergent.