Improper Integrals

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We study the concept of improper integrals.

(Video) Calculus: Single Variable

Online Texts

Examples

Compute the improper integral $\int _0^1 \ln x \, dx$ if it is convergent.

First off, we note that the integral is in fact improper because $\ln x$ has a vertical asymptote at $x = 0$ . We can find the antiderivative of $\ln x$ by integration by parts: $\int \ln x \, dx = \answer {x \ln x - x} + C$ (your answer should be the usual thing given by integration by parts. In particular, it equals $0$ at $x = e$ .)
The correct interpretation of the improper integral is $\int _0^1 \ln x \, dx = \lim _{b \rightarrow 0^+} \int _{\answer {b}}^{\answer {1}} \ln x \, dx.$
Using the antiderivative you found above (with $C = 0$ ), use the Fundamental Theorem of Calculus to conclude $\int _0^1 \ln x \, dx = \lim _{b \rightarrow 0^+} \int _{\answer {b}}^{\answer {1}} \ln x \, dx = \lim _{b \rightarrow 0^+} \left [ \left ( \answer {-1} \right ) - \left (\answer { b \ln b - b} \right ) \right ].$
As $b \rightarrow 0^+$ , $b \rightarrow 0$ and $\ln b \rightarrow - \infty$ . Using knowledge of orders of growth, we know that the product $b \ln b$
tends to $-\infty$ because $\ln b$ grows faster than any negative power of $b$ as $b \rightarrow 0^+$ . tends to some finite constant as $b \rightarrow 0^+$ because $\ln b$ has the same growth rate as $b^{-1}$ as $b \rightarrow 0^+$ . tends to zero because $\ln b$ grows slower than any negative power of $b$ as $b \rightarrow 0^+$ .
We take the limit as $b \rightarrow 0^+$ to conclude $\int _{0}^1 \ln x \, dx = \answer {-1}$ (write N/A if the limit diverges).

Without finding an antiderivative, determine whether the integral below is convergent or divergent. $\int _1^\infty \frac {dx}{x - \ln x}$

The dominant term in the denominator as $x \rightarrow \infty$ will be
$x$ $\ln x$
which suggests doing a comparison to the integral
$\displaystyle \int _1^\infty \frac {dx}{x}$ $\displaystyle \int _1^\infty \frac {dx}{- \ln x}$
Since $1/(x- \ln x)$ is always $1/x$ , the simplest test which applies is the direct comparisonlimit comparison test and it indicates convergencedivergencenothing–it does not apply .

Without finding an antiderivative, determine whether the integral below is convergent or divergent. $\int _0^1 \frac {dx}{2\sqrt {x} - x^2}.$

The dominant term in the denominator as $x \rightarrow 0^+$ is
$x^2$ $2 \sqrt {x}$
which suggests doing a comparison to the integral
$\displaystyle \int _0^1 \frac {dx}{2 \sqrt {x}}$ $\displaystyle \int _0^1 \frac {dx}{- x^2}$
Since $1/(2 \sqrt {x} -x^2)$ is always $1/(2 \sqrt {x})$ , the simplest test which applies is the direct comparisonlimit comparison test and it indicates convergencedivergencenothing–it does not apply .