Now this is an integral we can do easily:

$$ \begin{align} \lim_{t \rightarrow \infty} \, &\int_1^t \, \frac{1}{x} \, dx \\[5pt] &= \lim_{t \rightarrow \infty} \, ln|x| \bigg|_1^t \\[7pt] &= \lim_{t \rightarrow \infty} \, ln|t| - 0 \: \rightarrow \: \infty \end{align}$$

When the limit is evaluated, we see that the area under this curve is infinite. That's not too surprising given that we know that y = 0 is an asymptote, and that the function approaches but never quite reaches a value of zero as $x \rightarrow \infty.$ What's interesting is that similar functions do bound a finite area. We'll see one in the next example.

We say that this integral diverges. If one or both limits of an improper integral is infinite or doesn't exist, the integral diverges. An improper integral that has a finite limit is said to converge to that limit.

We say that an improper integral like this converges to a finite value. The interpretation of these two results that makes the most sense is that the second function approaches zero more rapidly, and apparently sufficiently fast that it encloses a finite area. You'll find a somewhat more satisfying interpretation in the section on the integral test for infinite series.

And below is a graph of these growing sums.

You can see that the sum of terms of the form 1/n² grows rapidly at first, then approaches a limit. In fact, the three digits after the decimal will settle in to .644 and never change thereafter. Further terms just refine digits successively to the right of the decimal. In contrast, the sum of terms of the form 1/n, the so-called harmonic series, continues to grow. In fact, it can be shown that this sum can be used to represent any number if we just add enough terms. That's the difference between divergence and convergence.

One way to think of it, if it helps, is that one denominator approaches infinity "faster" than the other, and that makes the difference.

This concept of convergence and divergence is central to many aspects of mathematics, including how computers and calculators compute things like sines, cosines and logarithms.

Now the fraction goes to zero as $t \rightarrow \infty$ as long as $p \gt 1,$ so as long as $p \gt 1,$ the integral has a finite value of $\frac{1}{p - 1}.$

This is a refinement in our understanding of the convergence of the area under functions like $f(x) = 1/x^p.$ As long as $p \gt 1,$ the area is finite and can be calculated.

Notice that for this integral, once we found the antiderivative, our discontinuity wasn't a problem any more, and simply performing the integral without regard to it would have worked. Nevertheless, it's important to be aware of discontinuities and make sure that any integral over one converges to a real value.

$$ \begin{align} \lim_{t \rightarrow 0^+} &[x \, ln(x) + x] \, \bigg|_t^1 \\[5pt] &= 1 \cdot ln(1) - 1 - \lim_{t \rightarrow 0^+} \, [t \, ln(t) - t] \\[5pt] &= -1 - \lim_{t \rightarrow 0^+} \, [t \, ln(t) - t] \\[5pt] &= -1 - \lim_{t \rightarrow 0^+} \, t \, ln(t) \end{align}$$

Now the remaining limit can be evaluated as an ∞/∞ limit using L'Hopital's rule:

$$ \begin{align} \lim_{t \rightarrow 0^+} \, t \, ln(t) &= \lim_{t \rightarrow 0^+} \, \frac{ln(t)}{1/t} \\[5pt] &= \frac{1/t}{-1/t^2} = -t \: \rightarrow \: 0 \end{align}$$

So remembering the -1 from our original limit, the result is -1. The shaded area of our integral is 1 square unit, lying below the x axis. The integral converges to a finite value.

Let's first think about this integral as if we didn't know about the discontinuity. It's pretty simple:

$$ \begin{align} \int_0^{10} \, \frac{1}{x - 2} \, dx &= ln |x - 2| \, \bigg|_0^{10} \\[5pt] &= ln|8| - ln|2| = ln \left| \frac{8}{2} \right| = ln(4) \end{align}$$

But this isn't correct, as we'll see. Now let's do the integral but this time consider the vertical asymptote at x = 2.

$$ \begin{align} &\lim_{t \rightarrow 2^-} \, int_0^t \, \frac{dx}{x - 2} + \lim_{t \rightarrow 2^+} \, int_t^{10} \, \frac{dx}{x - 2} \\[5pt] &= \lim_{t \rightarrow 2^-} ln|x - 2| \bigg|_0^t + \lim_{t \rightarrow 2^+} ln|x - 2| \bigg|_t^{10} \\[5pt] &= \lim_{t \rightarrow 2^-} ln|t - 2| - ln(2) \\[5pt] &\phantom{0000} + ln(8) - \lim_{t \rightarrow 2^+} ln|t - 2| \end{align}$$

Now both of those limits evaluate to ln(0), which is infinite, so this integral diverges, and the area it encloses is infinite. So you can see that we need to consider the discontinuity.

4. $$\int_{0}^5 \, \frac{x}{x - 1} \, dx$$

   Solution

   Here is the graph of our function:

   

   This is an improper integral because of the discontinuity at x = 1. Notice that if we just did the integral in a straightforward way, ignoring the vertical asymptote, we could do it by substitution: let $u = x - 1,$ then $du = dx$ and $x = u + 1,$ so we have

   $$ \begin{align} A &= \int_{-1}^4\, \frac{u + 1}{u} \, du \\[5pt] &= \int_{-1}^4\, 1 + \frac{1}{u} \, du \\[5pt] &= u + ln|u| \, \bigg|_{-1}^4 \\[5pt] &= 4 + ln(4) + 1 - ln|-1| = 5 + ln(4) \end{align}$$

   But this is not correct, as we shall see. The proper improper integral is

   $$A = \lim_{t \rightarrow 1} \, \int_0^t \, \frac{x}{x - 1} \, dx + \lim_{t \rightarrow 1} \, \int_t^5 \, \frac{x}{x - 1} \, dx$$

   Using the same substitution and swapping limits to u-limits, we have

   $$ \begin{align} &\lim_{t \rightarrow 1} \int_{-1}^{t-1} \left(1 + \frac{1}{u} \right) du + \lim_{t \rightarrow 1} \int_{t-1}^{4} \left(1 + \frac{1}{u} \right) du\\[5pt] &= \lim_{t \rightarrow 1} u + ln|u|\bigg|_{-1}^{t-1} + \lim_{t \rightarrow 1} u + ln|u| \bigg|_{t-1}^4 \end{align}$$

   This expression will contain two terms like

   $$\lim_{t \rightarrow 1} \, ln|t - 1|,$$

   so the limit is infinite, as is the area. The integral diverges.




5. $$\int_{0}^{\infty} \, x \, e^{-3x} \, dx$$

   Solution

   Here is the graph of our function:

   

   This is an improper integral because of the infinite upper limit. The improper integral is:

   $$\lim_{t \rightarrow \infty} \, \int_0^t x \, e^{-3x} \, dx$$

   We can do this integral by parts. Let $u = x,$ and $dv = e^{-3x} dx.$ Then $du = dx$ and $v = -\frac{1}{3} e^{-3x}.$ Then our integral becomes (forgetting the limits for now):

   $$\lim_{t \rightarrow \infty} \, -\frac{x}{3} e^{-3x} + \frac{1}{3} \int e^{-3x} \, dx$$

   Integrating and including the limits gives

   $$= \lim_{t \rightarrow \infty} \, -\frac{x}{3} e^{-3x} - \frac{1}{9} e^{-3x} \, \bigg|_0^t.$$

   Before evaluating the limits, it might be enlightening to rewrite that expression slightly, like this:

   $$= \lim_{t \rightarrow \infty} \, -\frac{x}{3 e^{3x}} - \frac{1}{9 e^{3x}} \, \bigg|_0^t.$$

   Now we can evaluate:

   $$\lim_{t \rightarrow \infty} \, \frac{-t}{3 e^{3t}} - \frac{1}{9 e^{3t}} + 0 + \frac{1}{9}$$

   The first limit can be evaluated using L'Hopital's rule. It reduces to zero. So our overal result is $\frac{1}{9}.$




6. $$\int_{-\infty}^{\infty} \, \frac{1}{1 + x^2} \, dx$$

   Solution

   Here is the graph of our function:

   

   This is an improper integral because of the infinite limits. Because of the symmetry of the function, we can just integrate from 0 to ∞ and multiply by 2. The improper integral is:

   $$2 \lim_{t \rightarrow \infty} \, \int_0^t \frac{1}{1 + x^2} \, dx$$

   This is an inverse tangent integral. Recall that $\frac{d}{dx} \text{tan}^{-1}(x) = \frac{1}{x^2 + 1},$ so we have

   $$2 \lim_{t \rightarrow \infty} \, \text{tan}^{-1}(x) \, \bigg|_0^t$$

   The inverse tangent of zero is zero, and the inverse tangent of ∞ is π/2, so our limit is

   $$= 2\lim_{t \rightarrow \infty} \, \text{tan}^{-1}(t) - \text{tan}^{-1}(0) = \pi \; units^2$$