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Section 7.5 Improper Integrals

An improper integral is an integral that either contians positive or negative infinity as point a or b, or contains a function that has a verticle asymptote somewhere between or at a or b.

Contrast the following functions:

Regular Integral

\begin{align*} \int _{1} ^{10} \dfrac{1}{x^2} dx \amp = \int _{1} ^{10} x^{-2} dx \\ \amp = \left[ \dfrac{x ^{-1} }{-1} \right] _{1} ^{10} \\ \amp = \left( - \dfrac{1}{10} \right) - \left( - \dfrac{1}{1} \right) \\ \amp = - \dfrac{1}{10} + 1\\ \amp = \dfrac{9}{10} \end{align*}

Note 7.5.1.

Notice: If we were to to go take the same integral from 1 to 100 we would get \(\dfrac{99}{100}.\)

Improper Integral

\begin{align*} \int _1 ^{\infty} \amp = \lim_{b\to\infty} \int_{1}^{b} \dfrac{1}{x^2} dx \\ \amp = \lim_{b\to\infty} \int_{1}^{b} x^{-2} dx \\ \amp = \lim_{b\to\infty} \left[ \dfrac{x^{-1}}{-1} \right] _{1}^{b} \\ \amp = \lim_{b\to\infty} \left( - \dfrac{1}{b} \right) - \left( - \dfrac{1}{1} \right) \\ \amp = \lim_{b\to\infty} - \dfrac{1}{b} + 1 \\ \amp = 0 + 1\\ \amp = 1 \end{align*}

Subsection 7.5.1 Improper Integrals to Infinity

The improper integral of f over the interval \([a\mbox{, }+\infty)\) is defined to be:

\begin{equation*} \int _{a}^{+\infty} f(x) dx = \lim_{b\to\infty} \int_{a}^{b} f(x) dx \end{equation*}

  • In the case where the limit exists, the improper integral converges and the limit if the value of the integral.

  • In the case where the limit doesn't exist, the improper integral diverges and it is not assigned a value.

The improper integral of f over the interval \(( -\infty\mbox{, }b]\) is defined to be:

\begin{equation*} \int_{-\infty}^{b} f(x) dx = \lim_{a\to - \infty} \int_{a}^{b} f(x) dx \end{equation*}

  • In the case where the limit exists, the improper integral converges and the limit if the value of the integral.

  • In the case where the limit doesn't exist, the improper integral diverges and it is not assigned a value.

Additionally,

The improper integral of f over the inverval \((-\infty\mbox{, }+\infty)\) is defined as:

\begin{equation*} \int_{-\infty}^{+\infty} f(x) dx = \int_{-\infty}^{c}f(x)dx + \int_{c}^{\infty}f(x)dx \end{equation*}

where c is any real number.

  • If both converge, then the functions converges.

  • Otherwise, the function diverges.

Subsection 7.5.2 Improper Integrals -- Asymptotes

We can also have improper integrals due to vertical asymptotes.

If f is continuous on \([a\mbox{, }b]\) except for an infinite discontinuity at b, then it is defined as:

\begin{equation*} \int _{a} ^{b} f(x) dx = \lim _{k \to b^{-}} \int _{a}^{k} f(x) dx \end{equation*}

or if there is an infinite discontinuity at a, then it is defined as:

\begin{equation*} \int _{a}^{b} f(x) dx = \lim _{k \to a^+} \int _{k}^{b} f(x) dx \end{equation*}
Example 7.5.1.

Evaluate: \(\displaystyle\int _0 ^8 \dfrac{1}{\sqrt[3]{x}} dx \)

Answer

\begin{align*} \int _0 ^8 \dfrac{1}{\sqrt[3]{x}} dx \amp = \lim _{k \to 0^{+}} \int _k ^8 \dfrac{1}{\sqrt[3]{x}} dx \\ \amp = \lim _{k \to 0^{+}} \int _k ^8 x ^{-\frac{1}{3}} dx \\ \amp = \lim _{k \to 0^{+}} \left[ \dfrac{x ^{\frac{2}{3}}}{\frac{2}{3}} \right] _k ^8 \\ \amp = \lim _{k \to 0^{+}} \left[ \dfrac{3}{2} x ^{\frac{2}{3}} \right] _k ^8 \\ \amp = \lim _{k \to 0^{+}} \left( \dfrac{3}{2} (8) ^{\frac{2}{3}} - \dfrac{3}{2} (k) ^{\frac{2}{3}} \right)\\ \amp = \dfrac{3}{2} (4) - \dfrac{3}{2} (0)\\ \amp = \dfrac{12}{2} - 0\\ \amp = 6 \end{align*}

Similarly to integrals to infinity, you can evaluate a function that has an asymptote between \([a\mbox{, }b]\) by splitting the function into two pieces.

If f is continuous on \([a\mbox{, }b]\) except for an infinite discontinuity at c in \((a\mbox{, }b\) is defined as:

\begin{equation*} \int _a ^b f(x) dx = \int _a ^c f(x) dx + \int _c ^ b f(x) dx \end{equation*}

  • If both converge, then the functions converges.

  • Otherwise, the function diverges.

Note 7.5.3.
You cannot simply apply the fundamental theorem of calculus directly to an improper integral. If you try you will get crazy results such as a negative area.