Improper Integrals

Evaluate the following improper integral:

$$\int_{1}^{\infty} \frac{1}{x^2} \mathrm{d}x.$$

Answer:

You should begin by noticing that, despite the function being discontinuous at \(x=0\), the discontinuity is not included in your interval of integration, so you are good to go.

Fig. 3. The discontinuity is not included in the integration interval.

You can now follow the steps for evaluating improper integrals.

1. Write the improper integral as a limit.

$$\int_{1}^{\infty} \frac{1}{x^2} \mathrm{d}x = \lim_{b\rightarrow\infty} \int_{1}^{b} \frac{1}{x^2} \mathrm{d}x.$$

2. Evaluate the definite integral.

For this step you can find the antiderivative of the function with the Power Rule,

$$\int \frac{1}{x^2} \mathrm{d}x = -\frac{1}{x}+C,$$

and then use the Fundamental Theorem of Calculus, so

$$\int_{1}^{\infty} \frac{1}{x^2} \mathrm{d}x=\lim_{b\rightarrow\infty} \left[ \left( -\frac{1}{b} \right) - \left( -\frac{1}{1} \right) \right].$$

3. Evaluate the limit and simplify.

$$ \begin{align} \int_{1}^{\infty}\frac{1}{x^2} \mathrm{d}x &= 0 -(-1) \\ &= 1. \end{align} $$

Evaluate the following integral:

$$\int_{-3}^{3}\frac{1}{x^3} \mathrm{d}x.$$

Answer:

1. Write both integrals using appropriate limits that approach the discontinuity.

Note that the function is discontinuous at \(x=0\), so you have to split it into two integrals

$$\int_{-3}^{3} \frac{1}{x^3} \mathrm{d}x = \int_{-3}^{0} \frac{1}{x^3} \mathrm{d}x + \int_{0}^{3} \frac{1}{x^3} \mathrm{d}x.$$

2. Evaluate the resulting integral(s).

Next, you need to evaluate each integral. This can be done by finding the antiderivative of the integrand (with the help of The Power Rule)

$$\int \frac{1}{x^3} \mathrm{d}x = -\frac{1}{2} \cdot \frac{1}{x^2} +C,$$

and then using The Fundamental Theorem of Calculus for each pair of integration limits, so

$$\int_{-3}^{0} \frac{1}{x^3} \mathrm{d}x=\lim_{t\rightarrow 0^{-}} \left[ \left( -\frac{1}{2} \cdot \frac{1}{t^2} \right) -\left(-\frac{1}{2} \cdot \frac{1}{(-3)^2}\right) \right],$$

and

$$\int_{0}^{3} \frac{1}{x^3} \mathrm{d}x=\lim_{t\rightarrow 0^{+}} \left[ \left( -\frac{1}{2} \cdot \frac{1}{3^2} \right) -\left(-\frac{1}{2} \cdot \frac{1}{t^2}\right) \right].$$

3. Evaluate the resulting limits.

You can note that neither of the required limits exist. This is enough information to conclude the integral diverges!

From our previous examples you found that

$$\int_{1}^{\infty} \frac{1}{x^2} \mathrm{d}x$$

converges, and its value is \(1\).

You also found that the value of

$$\int_{0}^{2} \frac{1}{x} \mathrm{d}x$$

does not exist because the limit involved in its evaluation does not exist, hence, you can say that it diverges.

Evaluate the following integral:

$$\int_{0}^{2} \frac{1}{x} \mathrm{d}x.$$

Answer:

Begin by noticing that the function \(f(x)=\frac{1}{x}\) is not defined when \(x=0\), so you are dealing with an improper integral.1. Write the improper integral as a limit.You should start by writing the improper integral as a limit, so

$$\int_{0}^{2} \frac{1}{x} \mathrm{d}x = \lim_{t\rightarrow 0^{+}} \int_{t}^{2} \frac{1}{x} \mathrm{d}x.$$

2. Evaluate the resulting integral.

Next, find the antiderivative of the function of the integrand (you can take a look at our Integrals Involving Logarithmic Functions for a refresher)

$$\int \frac{1}{x} \mathrm{d}x = \ln{x},$$

and finally, use The Fundamental Theorem of Calculus to evaluate the definite integral

$$\int_{0}^{2} \frac{1}{x} \mathrm{d}x = \ln{2} - \lim_{t\rightarrow 0^{+}} \ln{t}.$$

3. Evaluate the resulting limit.

Unfortunately, the last limit does not exist, so the integral does not exist either.

Evaluate the following integral:

$$\int_{0}^{\infty} e^{-x} \, \mathrm{d}x.$$

Answer:

1. Write the improper integral as a limit.

As usual, begin by writing the integral as a limit, so

$$\int_{0}^{\infty} e^{-x} \,\mathrm{d}x=\lim_{b\rightarrow \infty} \int_{0}^{b} e^{-x} \, \mathrm{d}x.$$

2. Evaluate the resulting integral.

Next, find the antiderivative of the exponential function

$$\int e^{-x} \,\mathrm{d}x = -e^{-x} + C.$$

Finally, you can use The Fundamental Theorem of Calculus to evaluate the definite integral

$$\int_{0}^{\infty} e^{-x} \,\mathrm{d}x = \lim_{b\rightarrow \infty} \left[ \left(-e^{-b} \right) - \left(-e^{0}\right) \right].$$

3. Evaluate the resulting limit.

The above limit of the exponential is equal to \(0\). Furthermore, any non-zero real number raised to the power of \(0\) is equal to \(1\). Therefore

$$\begin{align} \int_{0}^{\infty} e^{-x} \, \mathrm{d}x &= 0 - (-1) \\ &=1. \end{align}$$