Improper Integrals

Solved Problems

Solution.

We complete the square in the denominator:

Then

Hence, the integral converges.

Solution.

First we complete the square in the denominator:

Taking the limit, we obtain:

The integral converges.

Solution.

We can write this integral as

By the definition of an improper integral, we have

As both the limits exist and are finite, the given integral converges.

Solution.

This improper integral has an infinite upper limit of integration. Hence, by definition,

The integral converges and is equal to $$

Solution.

We can write the obvious inequality for the absolute values:

It's easy to show that the integral $$ converges (see also Example 1.) Indeed

Then we conclude that the integral $$ also converges by Comparison Test $$ Hence, the given integral $$ converges (absolutely) by Comparison Test $$

Solution.

There is a discontinuity in the integrand at $$ so that we must consider two improper integrals:

Using the definition of an improper integral, we obtain

For the first integral,

As it is divergent, the given integral $$ is also divergent.

Solution.

The integrand has discontinuity at the point $$ so that we can write the integral as

As you can see from this expression, there are $$ cases:

• If $$ then
  $$

  and the integral converges;

• If $$ then
  $$

  and the integral diverges.

Solution.

The given region is sketched in Figure $$

Since it is infinite, we calculate the improper integral to find the area:

Use integration by parts. Let $$ $$ Then $$ $$

Thus

We can apply L'Hopital's rule to find the limit:

Hence, the improper integral is

As you can see from the figure above, the required area is

Solution.

We calculate the length of the arc of the circle in the first quadrant between $$ and $$ and then multiply the result by $$

The equation of the circle centered at the origin is

Then the arc of the circle in the first quadrant (Figure $$) is described by the function

Find the derivative of the function:

Since the length of an arc is given by $$ we obtain

Now we calculate the improper integral $$

Thus, the circumference of the unit circle is $$