Define and calculate improper integrals with infinite limits of integration through examples with detailed solutions and explanations. Also more exercises with solutions are included.
Definition of Improper Integrals with Infinite Intervals [1] [2]
Consider 3 types of integral with infinite interval(s)
1 - The Upper Limit of the Integral is Infinite
An integral of the form \( \displaystyle \int_a^{\infty} f(x) \; dx \) exists and is convergent if \[ \displaystyle \int_a^{b} f(x) \; dx \quad \text{exists for all} \quad b \ge a \] and the limit
\[ \lim_{b \to \infty} \displaystyle \int_a^{b} f(x) \; dx \] exists and is finite.
We then write
\[ \boxed{ \displaystyle \int_a^{\infty} f(x) \; dx = \lim_{b \to \infty} \displaystyle \int_a^{b} f(x) \; dx } \]
2 - The Lower Limit of the Integral is Infinite
An integral of the form \( \displaystyle \int_{-\infty}^a f(x) \; dx \) exists and is convergent if \[ \displaystyle \int_b^{a} f(x) \; dx \quad \text{exists for all} \quad b \le a \] and the limit
\[ \lim_{b \to -\infty} \displaystyle \int_b^{a} f(x) \; dx \] exists and is finite.
We then write
\[ \boxed{ \displaystyle \int_{-\infty}^a f(x) \; dx = \lim_{b \to -\infty} \displaystyle \int_b^{a} f(x) } \]
3 - Both Limits of the Integral are Infinite
An integral of the form \( \displaystyle \int_{-\infty}^{\infty} f(x) \; dx \) exists and is convergent if both
\( \displaystyle \int_{-\infty}^a f(x) \; dx \) and \( \displaystyle \int_a^{\infty} f(x) \; dx \) are convergent
We then write
\[ \boxed{ \displaystyle \int_{-\infty}^{\infty} f(x) \; dx = \displaystyle \int_{-\infty}^a f(x) \; dx + \displaystyle \int_a^{\infty} f(x) \; dx } \]
Note that the interval of integration is split at \( x = a \) and hence we end up with the sum of two integrals with each integral having one infinite limit only.
Examples and Their Solutions
Example 1
Evaluate the integral given below if possible.
\[ \displaystyle \int_1^{\infty}\; x \; dx \]
Solution to Example 1
The given integral has the upper limit infinite and by definition in part (1) above, we write
\[ \displaystyle \int_1^{\infty}\; x \; dx = \lim_{b \to \infty} \int_1^{b}\; x \; dx \]
The limit \( \displaystyle \lim_{b \to \infty} (\dfrac{b^2}{2} - \dfrac{1}{2}) \) is infinite
Conclusion The limit is infinite and therefore the given improper integral \( \displaystyle \int_1^{\infty}\; x \; dx \) is divergent.
Example 2
Is the integral below convergent or divergent? Evaluate it if it is convergent.
\[ \displaystyle \int_1^{\infty}\; \dfrac{1}{x^2} \; dx \]
Solution to Example 2
The given integral has the upper limit infinite and by definition in part (1) above, we write
\[ \displaystyle \int_1^{\infty}\; \dfrac{1}{x^2} \; dx = \lim_{b \to \infty} \int_1^{b}\; \dfrac{1}{x^2} \; dx \]
limit \( \displaystyle \lim_{b \to \infty} (-\dfrac{1}{b} + 1) = 1 \) and hence
\[ \displaystyle \int_1^{\infty}\; \dfrac{1}{x^2} dx = 1 \]
Conclusion The given improper integral \( \displaystyle \int_1^{\infty}\; \dfrac{1}{x^2} \; dx \) is convergent.
Example 3
Evaluate the integral given below if possible.
\[ \displaystyle \int_{-\infty}^0\; \dfrac{1}{\sqrt{2-3x}} \; dx \]
Solution to Example 3
The given integral has the lower limit infinite and by definition in part (2) above, we write
\[ \displaystyle \int_{-\infty}^0 \; \dfrac{1}{\sqrt{2-3x}} \; dx = \lim_{b \to -\infty} \int_{b}^{0}\; \dfrac{1}{\sqrt{2-3x}} \; dx \]
Evaluate the above limit to obtain
\[ \displaystyle \lim_{b \to -\infty} \left(-\frac{2}{3}\sqrt{2} + \frac{2}{3}\sqrt{2-3b} \right) = -\frac{2}{3}\sqrt{2} + \infty = \infty \]
The limit is infinite and therefore given improper integral \( \displaystyle \int_{-\infty}^0\; \dfrac{1}{\sqrt{2-3x}} \; dx \) is divergent.
Example 4
Evaluate the integral given below if possible.
\[ \displaystyle \int_{-\infty}^{\infty}\; x^5 e^{-x^6} \; dx \]
Solution to Example 4
The given integral has both the lower and upper limits infinite and by definition in part (3) above, we split the integral as follows
\[ \displaystyle \int_{-\infty}^{\infty}\; x^5 e^{-x^6} \; dx = \int_{-\infty}^{0}\; x^5 e^{-x^6} \; dx + \int_{0}^{\infty}\; x^5 e^{-x^6} \; dx \qquad (I)\]
Note the limits of integration was split at \( x = 0 \) because this makes calculations easier. Other values might as well be selected to split the limit of integration and that would not change the result.
Evaluate the indefinite integral \( \displaystyle \int x^5 e^{-x^6} \; dx \) using the method of substitution
Let \( u = x^6 \) which gives \( \dfrac{du}{dx} = 6 x^5 \) which also gives \( dx = \dfrac{1}{6 x^5} \; du \); substitute and simplify to obtain
\( \displaystyle \int x^5 e^{-x^6} \; dx = \dfrac{1}{6} \int e^{-u} \; du = - \dfrac{1}{6} e^{-x^6} + c \) , where \( c \) is the constant of integration.
We now evaluate the two integrals on the right of (II) above
Example 5
For what values of \( p \) is the integral
\[ \displaystyle \int_{e}^{\infty}\; \dfrac{1}{x\:\left(\ln\:x\right)^{p+1}} \; dx \]
convergent? Find its value in terms of \( p \) .
Solution to Example 5
The given integral has the upper limit infinite and by definition in part (1) above, we write
\[ \displaystyle \int_{e}^{\infty}\; \dfrac{1}{x\:\left(\ln\:x\right)^{p+1}} \; dx = \lim_{b \to \infty} \int_e^{b}\; \dfrac{1}{x\:\left(\ln\:x\right)^{p+1}} \; dx \qquad (I)\]
Let \( u = \ln\:x \) , which gives \( \dfrac{du}{dx} = \dfrac{1}{x} \) and also \( dx = x \: du \)
Substitute and simplify the indefinite integral
\[ \int \dfrac{1}{x\:\left(\ln\:x\right)^{p+1}} \; dx = \int \dfrac{1} {u^{p+1}} \; du \]
Evaluate the integral on the right
\[ \int \dfrac{1}{x\:\left(\ln x \right)^{p+1}} \; dx = \int u^{-p-1} \; du = - \left(\dfrac{1}{p}\right) u^{-p} + c \]
where \( c \) is the constant of integration
Substitute back \( u = \ln\:x \)
\[ \int \dfrac{1}{x\:\left(\ln x \right)^{p+1}} \; du = - \dfrac{1}{p} (\ln x)^{-p} + c \]
Evaluate (I) above
\begin{align}
& \int_{e}^{\infty}\; \dfrac{1}{x\:\left(\ln\:x\right)^{p+1}} \; dx = \lim_{b \to \infty} \left[ - \dfrac{1}{p} (\ln x)^{-p} \right]_e^b \\[15pt]
& = \lim_{b \to \infty} \left[ - \dfrac{1}{p} (\ln b)^{-p} + \dfrac{1}{p} (\ln e)^{-p} \right] \\[15pt]
& = - \lim_{b \to \infty} \dfrac{1}{p} (\ln b)^{-p} + \dfrac{1}{p} (\ln e)^{-p} \\[15pt]
& \text{Simplify the term \( \dfrac{1}{p} (\ln e)^{-p} \) } \\[15pt]
& = - \lim_{b \to \infty} \dfrac{1}{p} (\ln b)^{-p} + \dfrac{1}{p}
\end{align}
For \( p =0 \), the above is undefined
For \( p \lt 0 \), \( \displaystyle - \lim_{b \to \infty} \dfrac{1}{p} (\ln b)^{-p} = \infty \)
For \( p \gt 0 \), \( \displaystyle - \lim_{b \to \infty} \dfrac{1}{p} (\ln b)^{-p} = 0 \)
Hence the given integral is convergent for \( p \gt 0 \) and is given by
\[ \boxed {\int_{e}^{\infty}\; \dfrac{1}{x\:\left(\ln\:x\right)^{p+1}} \; dx = \dfrac{1}{p} \quad \text{ for } p \; \gt 0 } \]
Exercises
Evaluate each of the following integrals if possible.
