Wemazi

Topic 10.4 · BC

Set up the related improper integral

Builds on checking the integral-test conditions. 8 minutes.

A region that never ends

The series \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n+1)^2}\) has the term \(\frac{1}{(2n+1)^2}\). Replace \(n\) by \(x\) to find the curve \(f(x)=\frac{1}{(2x+1)^2}\).

The series starts at \(n=1\), so the teal region starts at \(x=1\). The series has no last term, so the region has no right end. You stop at an orange edge \(b\) and find the area from 1 to \(b\).

Drag \(b\) to the right. At \(b=10\) the area is \(\frac17\), about 0.143, and at \(b=100\) it is about 0.164. The lower graph shows these areas on an axis that is squeezed, so that its last mark stands for infinity.

The area approaches \(\frac16\), about 0.167. This limit is the value of the improper integral \(\int_1^{\infty}f(x)\,dx\), an integral with an infinite upper limit.

The same two choices for every series

Without the numbers, the curve is \(f(x)\), the term of the series with \(x\) in place of \(n\).

Move the first index \(N\). The left edge of the region moves with it, because the region starts where the series starts.

Move \(b\) past the end of the figure. The region runs on toward infinity, because the series has no last term.

These are the only two choices: the term gives the function, and the first index gives the lower limit.

Notation

  1. Write the term \(a_n\) as a function \(f(x)\), so that \(f(n)=a_n\). The lower limit of the integral is the first index of the series.

    \[\sum_{n=N}^{\infty}a_n\;\longrightarrow\;\int_{\cy{N}}^{\infty}f(x)\,dx\]

  2. An improper integral is a limit. Integrate up to \(b\). Then let \(b\) grow.

    \[\int_N^{\infty}f(x)\,dx=\lim_{b\to\infty}\int_N^{b}f(x)\,dx\]

  3. Find an antiderivative \(F\). Substitute the limits. Then take the limit of \(F(b)\).

    \[\lim_{b\to\infty}\bigl(F(b)-F(N)\bigr)\]

  4. If the limit is a number, the integral converges. If the limit is infinite or does not exist, the integral diverges.

When it fails

Change the curve to \(f(x)=\dfrac{1}{2x+1}\). The area from 1 to \(b\) is \(\tfrac12\ln\dfrac{2b+1}{3}\), which is about 2.10 at \(b=100\) and 4.40 at \(b=10000\). The lower graph rises out of the top of the figure.

The area grows without end, so the integral diverges. The setup is the same as for the first curve, so only the limit of \(F(b)\) shows which result you have.

Example 1: A series that starts at 1

Set up and evaluate the improper integral for \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{(n^2+3)^2}\).

  1. Replace \(n\) by \(x\). The series starts at \(n=1\), so the lower limit is 1. Write the integral as a limit.

    \[\begin{aligned}&\int_1^{\infty}\frac{x}{(x^2+3)^2}\,dx\\ &=\lim_{b\to\infty}\int_1^{b}\frac{x}{(x^2+3)^2}\,dx\end{aligned}\]

  2. Substitute \(u=x^2+3\), so \(x\,dx=\tfrac12\,du\) and the integral becomes \(\tfrac12\int u^{-2}\,du=-\dfrac{1}{2u}\).

    \[\int\frac{x}{(x^2+3)^2}\,dx=-\frac{1}{2(x^2+3)}\]

  3. Substitute the limits 1 and \(b\).

    \[\begin{aligned}&\int_1^b\frac{x}{(x^2+3)^2}\,dx\\ &=\frac18-\frac{1}{2(b^2+3)}\end{aligned}\]

  4. As \(b\) grows, the last fraction goes to 0, so the improper integral converges.

    \[\lim_{b\to\infty}\left(\frac18-\frac{1}{2(b^2+3)}\right)=\cb{\tfrac18}\]

Example 2: A series that starts later

Set up and evaluate the improper integral for \(\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-1)^2}\).

  1. Replace \(n\) by \(x\) to find \(f(x)=\dfrac{1}{(x-1)^2}\).

  2. A tempting wrong move: Start the integral at 1, as in Example 1.

    \[\int_1^{\infty}\frac{1}{(x-1)^2}\,dx\]

    The bottom is 0 at \(x=1\), so the curve has a vertical asymptote there and \(f(1)\) does not exist. The first term of the series is \(a_3=\frac14\), so the region starts at \(x=3\).

  3. The first index is 3, so the lower limit is 3. An antiderivative is \(-\dfrac{1}{x-1}\).

    \[\int_3^{b}\frac{1}{(x-1)^2}\,dx=\frac12-\frac{1}{b-1}\]

  4. As \(b\) grows, \(\dfrac{1}{b-1}\) goes to 0, so the improper integral converges.

    \[\int_3^{\infty}\frac{1}{(x-1)^2}\,dx=\cb{\tfrac12}\]

Practice

Set up the improper integral for \(\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}\). Decide whether it converges.

The teal region starts at \(x=2\), where the series starts. Substitute \(u=\ln x\), and the integral becomes an integral of \(\frac1u\).

Solution

  1. Replace \(n\) by \(x\). The lower limit is 2.

    \[\begin{aligned}&\int_2^{\infty}\frac{1}{x\ln x}\,dx\\ &=\lim_{b\to\infty}\int_2^{b}\frac{1}{x\ln x}\,dx\end{aligned}\]

  2. Substitute \(u=\ln x\), so \(du=\dfrac{dx}{x}\) and the integral becomes \(\int\frac{du}{u}=\ln u\). An antiderivative is \(\ln(\ln x)\).

    \[\int_2^b\frac{1}{x\ln x}\,dx=\ln(\ln b)-\ln(\ln 2)\]

  3. As \(b\) grows, \(\ln b\) grows without bound, and so does \(\ln(\ln b)\). The limit is infinite, so the improper integral diverges.

    \[\lim_{b\to\infty}\ln(\ln b)=\infty\]

Questions

The graph shows \(f(x)=\dfrac{4}{(x+1)^2}\), the curve for the series \(\displaystyle\sum_{n=2}^{\infty}\frac{4}{(n+1)^2}\). The teal region ends at the orange edge \(b\), and the lower graph shows the area up to \(b\).

  1. At which value of \(x\) does the region start?

    • 2
    • 1
    • 0
    • 3
    Answer

    2

  2. At \(b=10\) the area is about 0.97, and the lower graph is still rising. What is \(\displaystyle\int_2^{\infty}f(x)\,dx\)?

    • \(\tfrac43\)
    • 0.97
    • It diverges.
    Answer

    \(\tfrac43\)

  3. Evaluate \(\displaystyle\int_1^{\infty}e^{-2x}\,dx\).

    • \(\dfrac{e^{-2}}{2}\)
    • \(e^{-2}\)
    • \(\tfrac12\)
    • It diverges.
    Answer

    \(\dfrac{e^{-2}}{2}\)

Summary

You replace \(n\) by \(x\) to find \(f\), start at the first index, and take the limit of the integrals up to \(b\).

This works because the region starts where the series starts, and the series has no last term.

This setup is the first step of the integral test for any series whose terms come from a positive, continuous, decreasing function.

This is one of four skills in topic 10.4 of Which & Why: AP Calculus.