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39. Improper Integrals1

Dated: 30-10-2024

Integrals over Infinite interval2

The improper integral1 is of the form

\[\int_a^{+\infty} f(x) dx\]

or

\[\int_{-\infty}^b f(x) dx\]

What Does it Mean?

\[\int_{a}^{+\infty}f(x)dx=\lim_{l\to+\infty}\int_{a}^{l}f(x)dx\]
  1. If the limit3 exists then we say the improper integral converges to a finite value.
  2. If the limit3 does not exist then we say the improper integral diverges and no finite value can be assigned.

Example

Evaluate \(\(\int_{1}^{+\infty}\frac{1}{x^{2}}dx\)\)

Solution

\[\int_{1}^{+\infty}\frac{1}{x^{2}}dx\]
\[=\lim_{l\to+\infty}\int_{1}^{l}\frac{1}{x^{2}}dx\]
\[=\lim_{l\to+\infty}\left(1-\frac{1}{l}\right)\]
\[=1-\frac{1}{\infty}=1-0=1\]

Example

Evaluate

\[\int_{1}^{+\infty}\frac{1}{x}dx\]

Solution

\[\int_{1}^{+\infty}\frac{1}{x}dx\]
\[= \lim_{l \to +\infty} \int_1^l \frac{1}{x} dx\]
\[= \lim_{l \to +\infty} \bigg[\ln \lvert x \rvert\bigg]_1^l\]
\[\lim_{l \to +\infty} \ln \lvert l \rvert = + \infty\]

If \(f(x)\) is not bound on interval2 then \(f(x)\) is not integratable over \([a, b]\).
We can get around this problem

  • If \(f(x)\) is continuous4 over the interval2 \([a, b)\) and does not have a limit3 from the left then we can define an improper integral in following way \(\(\int_a^b f(x) dx = \lim_{l \to b^-} \int_a^l f(x) dx\)\)

  • If \(f(x)\) is continuous4 over the interval2 \((a, b]\) and does not have a limit3 from the right then we can define an improper integral in following way \(\(\int_a^b f(x) dx = \lim_{l \to a^+} \int_l^b f(x) dx\)\)

If

  • \(f(x)\) is continuous4 over the interval2 \([a, b]\) except for the point \(c\)
  • \(a < c < b\)
  • \(f(x)\) goes infinite as \(x^+ \to c\) and \(x^- \to c\)
  • both improper integrals, which are \(\int_a^c f(x) dx\) and \(\int_c^b f(x) dx\), converge

then

\[\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx\]

Example

Evaluate

\[\int_{1}^{4}\frac{dx}{(x-2)^{2/3}}\]

Solution

The integrand1 approaches to \(+ \infty\) as \(x \to 2\). Therefore,

\[\int_{1}^{4}\frac{dx}{(x-2)^{2/3}} = \int_{1}^{2}\frac{dx}{(x-2)^{2/3}} + \int_{2}^{4}\frac{dx}{(x-2)^{2/3}}\]
\[\int_{1}^{2}\frac{dx}{(x-2)^{2/3}} \]
\[= \lim_{l\to 2^{-}}\int_{1}^{l}\frac{dx}{(x-2)^{2/3}} \]
\[= \lim_{l\to 2^{-}}\left[3(l-2)^{1/3}-3(1-2)^{1/3}\right] \]
\[= 3\]
\[\int_{2}^{4}\frac{dx}{(x-2)^{2/3}} \]
\[\lim_{l\to 2^{+}}\int_{l}^{4}\frac{dx}{(x-2)^{2/3}} \]
\[= \lim_{l\to 2^{+}}\left[3(4-2)^{1/3}-3(l-2)^{1/3}\right] \]
\[= 3\sqrt[3]{2}\]
\[\int_{1}^{4}\frac{dx}{(x-2)^{2/3}} = 3 + 3 \sqrt[3]{2}\]

References

Read more about notations and symbols.

  1. Read more about integrals. ↩↩↩

  2. Read more about intervals. ↩↩↩↩↩

  3. Read more about limits. ↩↩↩↩

  4. Read more about continuity. ↩↩↩