The integral in the figure below is evaluated by first performing a change a variable u = y^3, as indicated.

• We then express the improper integral as a limit in the usual manner.
• The limit of tan^-1(x) as x -> infinity is the angle approached as the tangent of the angle approaches infinity.
• The graph of the tangent function (second figure) shows that this angle is `pi/2.
• We thus obtain the values in the last line and the final result `pi/12.

The integral of dx / (x ln(x) ), shown below, is evaluated by the change of variable u = ln(x).

• After a few straightforward steps we obtain antiderivative ln( ln(x) ), to be evaluated at x = 2 and x = 1.
• We run into a serious problem when we try evaluate ln ( ln(1) ), since ln(1) = 0 and ln( ln(1) ) = ln(0) is undefined.
• This shows us that we have an improper integral.

We reevaluate the integral using the limit, as shown below.

• As y -> 1, ln(y) -> 0; thus the limit of ln ( ln(y) ), as y -> 1, is the limit of ln(z) as z -> 0. This limit is (-infinity).
• We thus see that the integral approaches infinity in the limit, and hence diverges.

In the figure below we evaluate the integrals of 1 / x^.9 and 1 / x^1.1, first from 1 to infinity then from 0 to 1.

The antiderivative of 1 / x^.9 is 10 x^.1.

At the top of the figure below we see a graph of the antiderivative 10 x^.1 of 1 / x^.9.

• The integral of 1 / x^.9 between two x values is the distance between the corresponding y values of the antiderivative function.
• The integral from 0 to 1 is therefore indicated by the vertical arrow from y = 0 to y = 10, as shown.
• The integral from x = 1 to infinity similarly corresponds to the y arrow from y = 10 to infinity.

In the lower half of the figure below we see the graph of the antiderivative -10 x^-.1 of 1 / x^1.1.

• The integrals from 1 to infinity and 0 to 1 are indicated by vertical arrows in a manner similar to that used in the top figure.

The above integrals show us why the integral of 1 / x^p, for x = 1 to infinity, diverges if p < 1 and converges if p > 1.

• Anytime p > 1, we end up with an antiderivative which is a power function with a negative power and which therefore approaches zero as x approaches infinity.
• Anytime p < 1, the antiderivative will be a power function with a positive power and will therefore approach infinity as x approaches infinity.

These integrals also show us why the integral of 1 / x^p from 0 to 1 must converge if p < 1 and diverge if p > 1.

The integral in the figure below corresponds to the indicated area under the graph.

The graph has a vertical asymptote at u = -4; its area might therefore converge or diverge, depending upon how tightly the asymptote is 'squeezed' to the vertical line x = -4.

• The change of variable t = u + 4, with the corresponding change of the limits of integration, shows us that we have the integral of dt / t^p for p < 1. Since the left-hand limit is 0, we see that the integral must therefore converge and the total area remains finite.

To test the convergence of the integral in the figure below, we need only note that the integrand is always < 1 / x^3, which converges.

• If the integrand is less than that of a convergent integral, the integral must converge.

Since 2^z = e^( z ln(2) ), 1/2^z = e^( - ln(2) z) = e^( -a z), with a = ln(2) > 0.

• Since the integral of e^(-a z) converges, so must the integral of 1 / 2^z.