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.

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.