{"id":774,"date":"2025-06-20T17:11:56","date_gmt":"2025-06-20T17:11:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=774"},"modified":"2025-07-22T13:58:55","modified_gmt":"2025-07-22T13:58:55","slug":"improper-integrals-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/improper-integrals-learn-it-2\/","title":{"raw":"Improper Integrals: Learn It 2","rendered":"Improper Integrals: Learn It 2"},"content":{"raw":"
Integrating a Discontinuous Integrand<\/h2>\r\n
So far we've dealt with infinite intervals. But what happens when the function itself has a problem\u2014like a vertical asymptote\u2014somewhere in our interval of integration?<\/p>\r\n
Consider [latex]\\int_a^b f(x)\u00a0 dx[\/latex] where [latex]f(x)[\/latex] is continuous on [latex][a,b)[\/latex] but has a discontinuity at [latex]x = b[\/latex]. Think about [latex]f(x) = \\frac{1}{\\sqrt{x-1}}[\/latex] on the interval [latex][1,2][\/latex]\u2014the function blows up as we approach [latex]x = 1[\/latex].<\/p>\r\n
Since [latex]f(x)[\/latex] is continuous on [latex][a,t][\/latex] for any [latex]t[\/latex] with [latex]a < t < b[\/latex], we can integrate from [latex]a[\/latex] to [latex]t[\/latex]. Then we see what happens as [latex]t[\/latex] approaches the discontinuity at [latex]b[\/latex].<\/p>\r\n
Figure 4 shows this visually\u2014as [latex]t[\/latex] gets closer to [latex]b[\/latex] from the left, we're asking whether the area under the curve approaches a finite value.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"] Figure 4. As [latex]t[\/latex] approaches b from the left, the value of the area from a to [latex]t[\/latex] approaches the area from a to b.[\/caption]<\/figure>\r\n\r\n
improper integrals with discontinuities<\/h3>\r\n
\r\n \t
Type 1: Discontinuity at the right endpoint<\/strong>\r\n
If [latex]f(x)[\/latex] is continuous over [latex][a,b)[\/latex]:<\/p>\r\n\r\n
Important<\/strong>: For Type 3, both<\/strong> integrals must converge for the whole integral to converge.<\/p>\r\n\r\n
\r\n \t
Convergence<\/strong>: The limit exists and is finite.<\/li>\r\n \t
Divergence<\/strong>: The limit doesn't exist or is infinite.<\/li>\r\n<\/ul>\r\n<\/section>\r\n
Let's see these definitions in action with some examples.<\/p>\r\n\r\n\r\n
\r\n
Evaluate [latex]{\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx[\/latex], if possible. State whether the integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n
\r\n
The function [latex]f\\left(x\\right)=\\frac{1}{\\sqrt{4-x}}[\/latex] is continuous over [latex]\\left[0,4\\right)[\/latex] and discontinuous at 4. Using equation 1 from the definition, rewrite [latex]{\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx[\/latex] as a limit:<\/p>\r\n\r\n
[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx& =\\underset{t\\to {4}^{-}}{\\text{lim}}{\\displaystyle\\int }_{0}^{t}\\frac{1}{\\sqrt{4-x}}dx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to {4}^{-}}{\\text{lim}}\\left(-2\\sqrt{4-x}\\right)|{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\underset{t\\to {4}^{-}}{\\text{lim}}\\left(-2\\sqrt{4-t}+4\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =4.\\hfill & & & \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n
The improper integral converges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
\r\n
Evaluate [latex]{\\displaystyle\\int }_{0}^{2}x\\text{ln}xdx[\/latex]. State whether the integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n
\r\n
Since [latex]f\\left(x\\right)=x\\ln{x}[\/latex] is continuous over [latex]\\left(0,2\\right][\/latex] and is discontinuous at zero, we can rewrite the integral in limit form using equation 2 from the definition:<\/p>\r\n\r\n
[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{0}^{2}x\\text{ln}xdx& =\\underset{t\\to {0}^{+}}{\\text{lim}}{\\displaystyle\\int }_{t}^{2}x\\text{ln}xdx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to {0}^{+}}{\\text{lim}}\\left(\\frac{1}{2}{x}^{2}\\text{ln}x-\\frac{1}{4}{x}^{2}\\right)|{}_{\\begin{array}{c}\\\\ t\\end{array}}^{\\begin{array}{c}2\\\\ \\end{array}}\\hfill & & & \\begin{array}{c}\\text{Evaluate}{\\displaystyle\\int}x\\ln{x}dx\\text{ using integration by parts}\\hfill \\\\ \\text{with }u=\\text{ln}x\\text{ and }dv=x.\\hfill \\end{array}\\hfill \\\\ & =\\underset{t\\to {0}^{+}}{\\text{lim}}\\left(2\\text{ln}2 - 1-\\frac{1}{2}{t}^{2}\\text{ln}t+\\frac{1}{4}{t}^{2}\\right).\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =2\\text{ln}2 - 1.\\hfill & & & \\begin{array}{c}\\text{Evaluate the limit.}\\underset{t\\to {0}^{+}}{\\text{lim}}{t}^{2}\\ln{t}\\text{ is indeterminate.}\\hfill \\\\ \\text{To evaluate it, rewrite as a quotient and apply}\\hfill \\\\ \\text{L'h}\\hat{o}\\text{pital's rule.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n
The improper integral converges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
\r\n
Evaluate [latex]{\\displaystyle\\int }_{-1}^{1}\\frac{1}{{x}^{3}}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n
\r\n
Since [latex]f\\left(x\\right)=\\frac{1}{{x}^{3}}[\/latex] is discontinuous at zero, using equation 3 from the definition, we can write<\/p>\r\n\r\n
If either of the two integrals diverges, then the original integral diverges. Begin with [latex]{\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx:[\/latex]<\/p>\r\n\r\n
[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }_{-1}^{0}\\frac{1}{{x}^{3}}dx& =\\underset{t\\to {0}^{-}}{\\text{lim}}{\\displaystyle\\int }_{-1}^{t}\\frac{1}{{x}^{3}}dx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to {0}^{-}}{\\text{lim}}\\left(-\\frac{1}{2{x}^{2}}\\right)|{}_{\\begin{array}{c}\\\\ -1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\underset{t\\to {0}^{-}}{\\text{lim}}\\left(-\\frac{1}{2{t}^{2}}+\\frac{1}{2}\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\text{+}\\infty .\\hfill & & & \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n
So far we’ve dealt with infinite intervals. But what happens when the function itself has a problem\u2014like a vertical asymptote\u2014somewhere in our interval of integration?<\/p>\n
Consider [latex]\\int_a^b f(x)\u00a0 dx[\/latex] where [latex]f(x)[\/latex] is continuous on [latex][a,b)[\/latex] but has a discontinuity at [latex]x = b[\/latex]. Think about [latex]f(x) = \\frac{1}{\\sqrt{x-1}}[\/latex] on the interval [latex][1,2][\/latex]\u2014the function blows up as we approach [latex]x = 1[\/latex].<\/p>\n
Since [latex]f(x)[\/latex] is continuous on [latex][a,t][\/latex] for any [latex]t[\/latex] with [latex]a < t < b[\/latex], we can integrate from [latex]a[\/latex] to [latex]t[\/latex]. Then we see what happens as [latex]t[\/latex] approaches the discontinuity at [latex]b[\/latex].<\/p>\n
Figure 4 shows this visually\u2014as [latex]t[\/latex] gets closer to [latex]b[\/latex] from the left, we’re asking whether the area under the curve approaches a finite value.<\/p>\n\nFigure 4. As [latex]t[\/latex] approaches b from the left, the value of the area from a to [latex]t[\/latex] approaches the area from a to b.<\/figcaption><\/figure>\n<\/figure>\n\n
improper integrals with discontinuities<\/h3>\n
\n
Type 1: Discontinuity at the right endpoint<\/strong>\n
If [latex]f(x)[\/latex] is continuous over [latex][a,b)[\/latex]:<\/p>\n
Important<\/strong>: For Type 3, both<\/strong> integrals must converge for the whole integral to converge.<\/p>\n
\n
Convergence<\/strong>: The limit exists and is finite.<\/li>\n
Divergence<\/strong>: The limit doesn’t exist or is infinite.<\/li>\n<\/ul>\n<\/section>\n
Let’s see these definitions in action with some examples.<\/p>\n\n
\n
Evaluate [latex]{\\displaystyle\\int }_{0}^{4}\\frac{1}{\\sqrt{4-x}}dx[\/latex], if possible. State whether the integral converges or diverges.<\/p>\n<\/div>\n
![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233856\/CNX_Calc_Figure_07_07_004.jpg\")
Figure 4. As [latex]t[\/latex] approaches b from the left, the value of the area from a to [latex]t[\/latex] approaches the area from a to b.[\/caption]<\/figure>\r\n