Power series offer a powerful solution when you encounter integrals that can’t be solved with standard techniques. You’ve already seen how power series help solve differential equations—now let’s explore their second major application: evaluating integrals involving functions whose antiderivatives don’t exist in elementary form.
Consider the integral [latex]\displaystyle\int e^{-x^2}dx[/latex], which appears frequently in probability theory and statistics. You might try various integration techniques, but you’ll quickly discover that none work. The reason? The antiderivative of [latex]e^{-x^2}[/latex] cannot be expressed as an elementary function – a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions.
The term “elementary” doesn’t mean “simple”—for instance, [latex]f(x) = \sqrt{x^2-3x} + e^{x^3} - \sin(5x+4)[/latex] is elementary despite its complexity.
When the antiderivative of a function isn’t elementary, we classify the integral as nonelementary.
nonelementary integrals
Any integral [latex]\displaystyle\int f(x)dx[/latex] where the antiderivative of [latex]f[/latex] cannot be written as an elementary function. These integrals require special techniques beyond basic integration methods.
When faced with nonelementary integrals, you can use this strategy: express the integrand as a power series, then integrate term by term. This approach transforms an impossible integral into a manageable infinite series.
We’ll demonstrate this technique using [latex]\displaystyle\int e^{-x^2}dx[/latex] as our key example.
Express [latex]\displaystyle\int {e}^{\text{-}{x}^{2}}dx[/latex] as an infinite series.
Evaluate [latex]{\displaystyle\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx[/latex] to within an error of [latex]0.01[/latex].
The Maclaurin series for [latex]{e}^{\text{-}{x}^{2}}[/latex] is given by
The sum of the first four terms is approximately [latex]0.74[/latex]. By the alternating series test, this estimate is accurate to within an error of less than [latex]\frac{1}{216}\approx 0.0046296<0.01[/latex].
Watch the following video to see the worked solution to the example above.
The integral [latex]\displaystyle\int e^{-x^2}dx[/latex] isn’t just a mathematical curiosity—it plays a crucial role in analyzing real-world data. You’ll encounter this integral when working with normally distributed data sets, where values follow the familiar bell-shaped curve pattern.
When data is normally distributed with mean [latex]\mu[/latex] and standard deviation [latex]\sigma[/latex], you can find the probability that a randomly chosen value falls between [latex]x = a[/latex] and [latex]x = b[/latex] using:
This represents the area under the normal curve between the two boundary values. (See Figure 2.)
Figure 2. If data values are normally distributed with mean [latex]\mu [/latex] and standard deviation [latex]\sigma [/latex], the probability that a randomly selected data value is between [latex]a[/latex] and [latex]b[/latex] is the area under the curve [latex]y=\frac{1}{\sigma\sqrt{2\pi}}{e}^{\frac{-\left(x-\mu\right)^{2}}{\left(2{\sigma}^{2}\right)}}[/latex] between [latex]x=a[/latex] and [latex]x=b[/latex].
To simplify calculations, statisticians use the substitution [latex]z = \frac{x-\mu}{\sigma}[/latex]. This quantity [latex]z[/latex] is called the z-score of a data value—it measures how many standard deviations away from the mean a particular value lies.
With this substitution, our probability integral becomes the more manageable form:
This standardized version allows us to work with universal probability tables and makes calculations more efficient. In the next example, we’ll show how to use this integral to calculate actual probabilities.
Suppose a set of standardized test scores are normally distributed with mean [latex]\mu =100[/latex] and standard deviation [latex]\sigma =50[/latex]. Use the equation before this example and the first six terms in the Maclaurin series for [latex]{e}^{\frac{\text{-}{x}^{2}}{2}}[/latex] to approximate the probability that a randomly selected test score is between [latex]x=100[/latex] and [latex]x=200[/latex]. Use the alternating series test to determine how accurate your approximation is.
Since [latex]\mu =100,\sigma =50[/latex], and we are trying to determine the area under the curve from [latex]a=100[/latex] to [latex]b=200[/latex], the integral becomes
Using the first five terms, we estimate that the probability is approximately [latex]0.4922[/latex]. By the alternating series test, we see that this estimate is accurate to within
If you are familiar with probability theory, you may know that the probability that a data value is within two standard deviations of the mean is approximately [latex]95\%[/latex]. Here we calculated the probability that a data value is between the mean and two standard deviations above the mean, so the estimate should be around [latex]47.5\text{%}[/latex]. The estimate, combined with the bound on the accuracy, falls within this range.
Nonelementary integrals appear in many areas of physics and engineering. One important example comes from analyzing pendulum motion, where the period involves the integral:
This integral cannot be evaluated using standard techniques. An integral of this form is known as an elliptic integral of the first kind. These integrals originally arose when mathematicians attempted to calculate the arc length of an ellipse, which explains their name. Elliptic integrals appear throughout physics and engineering—from pendulum periods to planetary motion calculations. Since we can’t solve them analytically, power series provide an essential tool for obtaining numerical approximations.
We’ll now demonstrate how to use power series techniques to approximate this integral and make it useful for practical calculations.
The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length [latex]L[/latex] that makes a maximum angle [latex]{\theta }_{\text{max}}[/latex] with the vertical, its period [latex]T[/latex] is given by
where [latex]g[/latex] is the acceleration due to gravity and [latex]k=\sin\left(\frac{{\theta }_{\text{max}}}{2}\right)[/latex] (see Figure 3). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and [latex]\sin\theta[/latex] is approximated by [latex]\theta[/latex].) Use the binomial series
to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if
you use only the first term in the binomial series, and
you use the first two terms in the binomial series. Figure 3. This pendulum has length [latex]L[/latex] and makes a maximum angle [latex]{\theta }_{\text{max}}[/latex] with the vertical.
We use the binomial series, replacing [latex]x[/latex] with [latex]\text{-}{k}^{2}{\sin}^{2}\theta[/latex]. Then we can write the period as
If [latex]{\theta }_{\text{max}}[/latex] is small, then [latex]k=\sin\left(\frac{{\theta }_{\text{max}}}{2}\right)[/latex] is small. We claim that when [latex]k[/latex] is small, this is a good estimate. To justify this claim, consider
Furthermore, it can be shown that each coefficient on the right-hand side is less than [latex]1[/latex] and, therefore, that this expression is bounded by
For larger values of [latex]{\theta }_{\text{max}}[/latex], we can approximate [latex]T[/latex] by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate
The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are useful because they allow us to represent known functions using polynomials, thus providing us a tool for approximating function values and estimating complicated integrals. In addition, they allow us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.
