We interpret definite integrals as giving us the net area underneath the graph of the function over the given interval. If we used this interpretation here, this notation should mean that we want to find the area under \(\frac {1}{x}\) from \(1\) to \(\infty \). This idea needs to be more precisely defined. Let’s consider the definite integral

where \(b \geq 1\) is some fixed number.

This integral gives us the area underneath \(1/x\) from \(1\) to \(b\). If we wanted to make sense of the integral to \(\infty \), we could think of letting \(b\) get larger and larger and seeing how this affects the area.

That is, we interpret the integral from 1 to \(\infty \) as a limit of a definite integral:

We can interpret this result to mean that the area under \(\frac {1}{x}\) from \(1\) to \(\infty \) is infinite.

The previous example shows us we should consider this expression to be the limit of a definite integral in the following manner:

We calculate the definite integral

This gives the area under \(\frac {1}{x^2}\) on the interval \([1, b]\).

Now we take the limit of this area as \(b \to \infty \).

In this case, when we take the limit, we get a finite number, namely \(1\). This suggests that we can consider the area under \(1/x^2\) from \(1\) to \(\infty \) to be equal to \(1\) despite the fact that the region is unbounded.

Here both of the bounds of integration involve \(\infty \) so to reinterpret the integral in terms of limits of definite integrals, we must first break up the integral into two pieces.

\begin{align*} \int _{-\infty }^{\infty } \frac {1}{1+x^2} dx&= \int _{-\infty }^{0} \frac {1}{1+x^2} dx + \int _{0}^{\infty } \frac {1}{1+x^2} dx \\ & =\lim _{a \to -\infty } \int _{a}^{0} \frac {1}{1+x^2} dx + \lim _{b \to \infty } \int _{0}^{\infty } \frac {1}{1+x^2} dx \end{align*}

Now we must look at each integral separately.

(your answer should be in terms of \(b\)) Similarly we obtain (your answer should be in terms of \(a\))

Now we take the limits of each integral

Hence we have

In this example, both integrals converge and therefore the entire integral converges to \(\pi \).

Write with me, \begin{align*} \int _{-\infty }^{-1} \frac {1}{x\sqrt {x^2-1}} dx &= \lim _{a \to -\infty } \int _{a}^{-1} \frac {-1}{x\sqrt {x^2-1}}dx\\ &= \lim _{a \to -\infty } \bigg [arcsec(x)\bigg ]_{a}^{-1} \\ &= \lim _{a \to -\infty } \left (arcsec(-1) - arcsec(a)\right )\\ &= \answer [given]{\pi - \frac {\pi }{2}}. \end{align*}

Write with me: Let us evaluate these limits one at a time: \begin{align*} \lim _{a\to -\infty }\bigg [-\cos (\theta )\bigg ]_a^1 &= \lim _{a\to -\infty }\left (-\cos (1)+\cos (a)\right )\\ &= \lim _{a\to -\infty } \cos (a) -\cos (1)\\ &= \answer [given]{DNE}. \end{align*}

We can stop here. Since one limit did not exist, the integral is divergent, and hence does not exist (DNE). We do not need to evaluate the other integral.

It may not be obvious at first glance that this integral is improper, but it is. Here the problem is that the function \(\frac {1}{\sqrt {x}}\) is unbounded on \([0,1]\) because \(\frac {1}{\sqrt {x}}\) gets arbitrarily large as \(x\) approaches \(0\) from the right That is, there is a vertical asymptote at \(x=0\) because \(\lim _{x \to 0^+} \frac {1}{\sqrt {x}} dx = \infty \). From now on, we will need to be cautious when evaluating integrals to check whether the integrand is bounded on the region of integration.

How then do we work with integrals of unbounded functions? Can we define this integral in terms of integrals we know how to compute?

Suppose \(b\) lies strictly between \(0\) and \(1\). Then consider the definite integral

The function is now bounded on this smaller interval.

Now we can examine what will happen if we let \(b\) begin to move towards \(0\) from the right. What will happen to the area under the curve? Just as we did in the case of unbounded integrals, we can make sense of this integral by defining it as the limit of a definite integral

We can then evaluate our original improper integral: \begin{align*} \int _{0}^1 \frac {1}{\sqrt {x}} dx &= \lim _{\answer [given]{b} \to 0} \int _{b}^1 \frac {1}{\sqrt {x}} dx \\ &= \lim _{b \to 0} \bigg [\answer [given]{2\sqrt {x}}\bigg ]_b^1 \\ &= \lim _{b \to 0} \left (\answer [given]{2-2\sqrt {b}}\right )\\ &= \answer [given]{2}. \end{align*}

Isn’t it quite surprising that the area bounded by a curve which has a vertical asymptote can be finite? Consider the graph of \(y= \frac {1}{\sqrt {x}}\) on the interval \([0,1]\):

Essentially, the function \(\frac {1}{\sqrt {x}}\) approaches the vertical asymptote at \(x=0\) “fast enough” so that only a finite amount of area is accumulated.

The function \(\frac {1}{x^2}\) has a vertical asymptote at \(a=0\). Since the problematic point occurs in the middle of the interval we need to break the integral into two pieces. We have \begin{align*} \int _{-1}^1 \frac {1}{x^2} dx =&= \int _{-1}^{0} \frac {1}{x^2} dx + \int _{0}^{1} \frac {1}{x^2} dx \\ &= \lim _{a \to 0^+} \int _{a}^1 \frac {1}{x^2} dx + \lim _{b \to 0^-} \int _{-1}^{b} \frac {1}{x^2} dx. \end{align*}

Let us evaluate these limits one at a time. \begin{align*} \lim _{a \to 0^+} \int _{a}^1 \frac {1}{x^2} dx &= \lim _{\answer [given]{a} \to 0^+} \bigg [\answer [given]{\frac {-1}{x}}\bigg ]_a^1 \\ &= \lim _{a \to 0^+} \frac {1}{a}-1\\ &= \infty . \end{align*}

We can stop here. Since one limit was infinite, the integral is divergent, and hence does not exist (DNE).

Since \(\lim _{x\to 0^+} \ln (x) = -\infty \), the natural logarithm has a vertical asymptote at \(x = 0\). Thus, we have \begin{align*} \int _0^1 \ln (x) dx &=\lim _{\answer [given]{t}\to 0^+} \int _t^1 \ln (x) dx \\ &= \lim _{t\to 0^+} \bigg [\answer [given]{x \ln (x)-x}\bigg ]_t^1 \\ &=\lim _{t \to 0^{+}} ( -1 + t - t\ln (t) ) \\ &=\answer [given]{ -1} \end{align*}

This is clearly an improper integral because one of the bounds is infinite, but we need to be cautious about the integrand as well. The integrand is a rational function so we check whether the denominator has a root in \([1,\infty )\). Factoring the denominator \(x^2+x-6=(x-2)(x+3)\) so we see that \(x=2\) makes the denominator zero and lies in the interval of integration.

Since \(2\) lies in the middle of the interval and the upper limit of integration is infinity, we need to break the integral into three pieces.

\begin{align*} & \int _{1}^{\infty } \frac {5x}{x^2+x-6} dx \\ &= \int _{1}^{2} \frac {5x}{(x-2)(x+3)} dx + \int _{2}^{4} \frac {5x}{(x-2)(x+3)} dx + \int _{4}^{\infty } \frac {5x}{(x-2)(x+3)} dx\\ \end{align*}

We chose \(4\) for the upper and lower limit of the middle and last integral, respectively, but we could use any number greater than \(2\).

\begin{align*} & \int _{1}^{\infty } \frac {5x}{x^2+x-6} dx \\ &=\lim _{a \to 2^{-}} \int _{1}^a \frac {5x}{(x-2)(x+3)} dx + \lim _{b \to 2^{+}} \int _{b}^{4} \frac {5x}{(x-2)(x+3)} dx +\lim _{c \to \infty } \int _{4}^{c} \frac {5x}{(x-2)(x+3)} dx \end{align*}

Since the integrand is a rational function, where the denominator is reducible, we could apply partial fractions decomposition to get

Now we find the antiderivative to get \begin{align*} \int \frac {5x}{(x-2)(x+3)} dx&=\int \frac { 2}{x-2} dx + \int \frac { 3}{x+3} dx \\ =& 2\ln |x-2| + 3\ln |x+3 | + C \end{align*}

Now we evaluate one integral at a time. Let’s start with

\begin{align*} \lim _{b \to 2^{+}} \int _{b}^{4} \frac {5x}{(x-2)(x+3)} dx&= \lim _{b \to 2^{+}} \bigg [ 2\ln |x-2| + 3\ln |x+3 | \bigg ]_{b}^{4} \\ &=\lim _{b \to 2^{+}} (2\ln (2) + 3 \ln (7) - 2\ln |b-2| -3\ln |b+3| ) \\ &=2\ln (2) +3\ln (7)-3\ln (5) + \infty =\infty \end{align*}

Thus this integral diverges, so the entire integral must diverge.