Welcome to an example of an improper integral,
where we have an infinite interval.
We will determine if the given integral
converges or diverges, and if it converges,
we want to evaluate the integral.
So, looking at the integral, notice how
the lower limit of integration is negative infinity,
and the upper limit of integration is positive infinity.
So, for a quick review,
because we have an infinite interval,
on the interval from negative infinity to positive infinity,
we're first going to write the integral
as two separate integrals.
By breaking it up using a constant c,
any constant between negative infinity
and positive infinity.
And then, once we have these two separate integrals,
we'll write them using this limit notation, as we see above.
And because the domain of our function is all real numbers,
let's go ahead and break this up into separate integrals
using the constant, zero.
Which means, we'll first have the integral from
negative infinity to zero,
plus the integral from zero to positive infinity.
And now, we're going to rewrite this using limits.
We'll write this first integral as the limit
as c approaches a negative infinity
of the integral from c to zero,
plus the limit as c approaches positive infinity
of the integral from zero to c.
Now, we'll integrate, and determine the limits.
So, for the integral of one
divided by the quantity one plus x squared.
To determine the integral,
we'll be using this integral formula here.
Notice, a would be one, and u is just x.
So, we'll have the limit
as c approaches negative infinity
of arctangent x
and we still have our limits of integration from c to zero,
plus the limit as c approaches positive infinity
of arctangent x.
The limits of integration are from zero to c,
which means we'll have the limit
as c approaches a negative infinity
of arctangent zero,
minus arctangent c,
plus the limit as c approaches positive infinity
of arctangent c
minus arctangent zero.
And now, to find these limits,
notice that arctangent zero is not affected by c,
so for arctangent zero, we're looking for an angle
that has a tangent function value of zero.
So, if we look at the graph of the tangent function,
notice how the tangent function value is zero,
where theta, the angle, is zero.
So, here we'd have zero minus,
now c approaches negative infinity,
we're looking for an angle
such that the tangent function value
would be approaching negative infinity.
So again, going back to our graph,
notice how the tangent function value
is approaching negative infinity, as our angle, theta,
is approaching negative pi over two.
So, we'd have minus negative pi over two,
plus, now c's approaching positive infinity,
so for arctangent c,
where c is approaching positive infinity,
we're looking for an angle, theta,
such that the tangent function value
is now approaching positive infinity.
So, going back to our graph again, of the tangent function,
notice that tangent function values
are approaching positive infinity,
when our angle, theta, is approaching pi over two.
So, we'd have plus pi over two, minus,
arctangent zero is zero.
Notice how this would be a positive pi over two
plus positive pi over two,
which would be two pi over two, or just pi.
So, because the limit exists, and is equal to pi,
this is also the value of our def integral,
and therefore, we can say the improper integral converges.
And also because the integrand, or the function,
one divided by the quantity one plus x squared
is nonnegative on this infinite interval,
we can say that the area under this function,
and above the x axis, even though it's unbounded,
would be equal to pi square units.
I hope you found this explanation helpful.