And now we will do a very funny thing.
We will integrate to infinity.
Let’s calculate, for instance, this area:
We will integrate to by first integrating to ,
and then we say “Dear , please go to “.
Let’s see what this limit would be.
These may be handy for the calculation:
But it is easier to remember them like this.
Here comes another one:
Well, it could happen that both limits are infinite:
In that case, we split the integration, let’s say, at zero.
Actually, this area is 1/3+1/3, so 2/3, but definite integration works in a way that areas under the x axis will get a negative sign.
So that’s why we got zero.
These integrals that stretch into infinity are called improper integrals.
Those that we saw so far were stretching into infinity in the x axis direction, but there are others that do the same along the y axis.
Here it is:
But the solution process is the same.
If we integrate this function over the positive number line:
then we get an improper integral from 0 to 1,
and also from 1 to infinity.
Let’s see first what happens if we integrate from 0 to 1.
Well, we will discuss the case of separately.
Now let's see what this limit is.
First we substitute 1,
and then see what happens if .
That means if the exponent is a positive number,
then we will get zero here.
If the exponent is negative...
then the limit is infinite.
If happens to be 1:
Now let’s see what is going on from 1 to infinity.
If , then the exponent is positive,
and in that case the integral is divergent.
And if happens to be 1:
Summarizing all this: if , then from 0 to 1 the integral is divergent, and from 1 to infinity it is convergent.
If , then from 0 to 1 the integral is convergent, and from 1 to infinity it is divergent.
And if , well, then it is divergent everywhere.