Improper Integral

Tags

Calculus

Cegep/2

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903 words

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6 minutes

Extension of definite integral to esoteric cases where FTC2 doesn't apply

Forms

Infinite integral

If $f$ is continuous on $[a,\mathrm{\infty })$, then

$${\int }_a^{\mathrm{\infty }}f(x)\mathrm{d}x=\underset{t\to \mathrm{\infty }}{lim}{\int }_a^{t}f(x)\mathrm{d}x$$

Discontinuous integral

If $f$ is discontinuous at $a$, then

$${\int }_a^{b}f(x)\mathrm{d}x=\underset{t\to a^{+}}{lim}{\int }_t^{b}f(x)\mathrm{d}x$$

If the discontinuity occurs inside the interval, split the integral.

Examples

Evaluate the following infinite integrals:

${\int }_{\ln 2}^{\mathrm{\infty }}\frac{e^x}{(e^{2x}+4{)}^{\frac{3}{2}}}\mathrm{d}x$

$${\int }_{\ln 2}^{\mathrm{\infty }}\frac{e^x}{(e^{2x}+4{)}^{\frac{3}{2}}}\mathrm{d}x=\underset{t\to \mathrm{\infty }}{lim}{\int }_{\ln 2}^t\frac{e^x}{(e^{2x}+4{)}^{\frac{3}{2}}}\mathrm{d}x$$

Let $u=e^x$, then $du=e^{x}dx$.

$$=\underset{t\to \mathrm{\infty }}{lim}{\int }_2^{e^t}\frac{1}{(u^2+4{)}^{\frac{3}{2}}}\mathrm{d}u$$

Let $u=2\tan \theta$.

$$\begin{array}{rl}& =\underset{t\to \mathrm{\infty }}{lim}{\int }_{\frac{\pi }{4}}^{\arctan \frac{e^t}{2}}\frac{2{\sec }^2\theta }{(2\sec \theta {)}^3}\mathrm{d}\theta \\ & =\underset{t\to \mathrm{\infty }}{lim}\frac{1}{4}{\int }_{\frac{\pi }{4}}^{\arctan \frac{e^t}{2}}\cos \theta \mathrm{d}\theta \\ & =\underset{t\to \mathrm{\infty }}{lim}{\frac{1}{4}\sin \theta |}_{\frac{\pi }{4}}^{\arctan \frac{e^t}{2}}\\ & =\underset{t\to \mathrm{\infty }}{lim}\frac{1}{4}[\sin (\arctan \frac{e^t}{2})-\sin \frac{\pi }{4}]\\ & =\frac{1}{4}[\sin \frac{\pi }{2}-\sin \frac{\pi }{4}]\\ & =\frac{1}{4}(1-\frac{\sqrt{2}}{2})\end{array}$$

Evaluate the following discontinuous integrals:

${\int }_{-1}^1\frac{1}{x^2}\mathrm{d}x$

$$\begin{array}{rl}{\int }_{-1}^1\frac{1}{x^2}\mathrm{d}x& =2{\int }_0^1\frac{1}{x^2}\mathrm{d}x\\ & =2\underset{t\to 0^{+}}{lim}{\int }_t^1\frac{1}{x^2}\mathrm{d}x\\ & =2{\underset{t\to 0^{+}}{lim}(-\frac{1}{x})|}_t^1\\ & =2\underset{t\to 0^{+}}{lim}(-\frac{1}{1}+\frac{1}{t})\\ & =\mathrm{\infty }\end{array}$$

${\int }_1^4\frac{1}{x-\sqrt{x}}\mathrm{d}x$

$${\int }_1^4\frac{1}{x-\sqrt{x}}\mathrm{d}x=\underset{t\to 1^{+}}{lim}{\int }_t^4\frac{1}{x-\sqrt{x}}\mathrm{d}x$$

Let $u=\sqrt{x}$, then $du=\frac{1}{2\sqrt{x}}dx$.

$$\begin{array}{rl}& =\underset{t\to 1^{+}}{lim}{\int }_{\sqrt{t}}^2\frac{2u}{u^2-u}\mathrm{d}u\\ & =\underset{t\to 1^{+}}{lim}2{\int }_{\sqrt{t}}^2\frac{1}{u-1}\mathrm{d}x\\ & =\underset{t\to 1^{+}}{lim}2{\ln |u-1||}_{\sqrt{t}}^2\\ & =\underset{t\to 1^{+}}{lim}2(\ln |2-1|-\ln |\sqrt{t}-1|)\\ & =\mathrm{\infty }\end{array}$$

${\int }_2^3\frac{1}{\sqrt{3-x}}\mathrm{d}x$

$${\int }_2^3\frac{1}{\sqrt{3-x}}\mathrm{d}x=\underset{t\to 3^{-}}{lim}{\int }_2^t\frac{1}{\sqrt{3-x}}\mathrm{d}x$$

Let $u=3-x$, then $du=-dx$.

$$\begin{array}{rl}& =\underset{t\to 3^{-}}{lim}-{\int }_1^{3-t}\frac{1}{\sqrt{u}}\mathrm{d}u\\ & =\underset{t\to 3^{-}}{lim}-[2\sqrt{u}{]}_1^{3-t}\\ & =\underset{t\to 3^{-}}{lim}-(2\sqrt{3-t}-2\sqrt{1})\\ & =2\end{array}$$

${\int }_0^1\sqrt{x}\ln x\mathrm{d}x$

$${\int }_0^1\sqrt{x}\ln x\mathrm{d}x=\underset{t\to 0^{+}}{lim}{\int }_t^1\sqrt{x}\ln x\mathrm{d}x$$

Let $u=\ln x$ and $dv=\sqrt{x}dx$,
then $du=\frac{1}{x}dx$ and $v=\frac{2}{3}{x}^{\frac{3}{2}}$.

$$\begin{array}{rl}& =\underset{t\to 0^{+}}{lim}{\ln x\cdot \frac{2}{3}{x}^{\frac{3}{2}}-\int \frac{2}{3}{x}^{\frac{1}{2}}\mathrm{d}u|}_t^1\\ & =\underset{t\to 0^{+}}{lim}{\frac{2}{3}\ln x\cdot x^{\frac{3}{2}}-\frac{4}{9}{x}^{\frac{3}{2}}|}_t^1\\ & =\underset{t\to 0^{+}}{lim}(\frac{2}{3}\ln 1\cdot 1^{\frac{3}{2}}-\frac{4}{9}{1}^{\frac{3}{2}})-(\frac{2}{3}\ln t\cdot t^{\frac{3}{2}}-\frac{4}{9}{t}^{\frac{3}{2}})\\ & =-\frac{4}{9}-\frac{2}{3}\underset{t\to 0^{+}}{lim}\ln t\cdot t^{\frac{3}{2}}\\ & =-\frac{4}{9}-\frac{2}{3}\underset{t\to 0^{+}}{lim}\frac{\ln t}{t^{-\frac{3}{2}}}\\ & =-\frac{4}{9}-\frac{2}{3}\underset{t\to 0^{+}}{lim}\frac{\frac{1}{t}}{-\frac{3}{2}{t}^{-\frac{5}{2}}}\\ & =-\frac{4}{9}-\frac{2}{3}\underset{t\to 0^{+}}{lim}-\frac{2}{3}{t}^{\frac{3}{2}}\\ & =-\frac{4}{9}\end{array}$$

Determine if the improper integral converges or diverges. If it converges, find the value:

${\int }_0^{\mathrm{\infty }}\frac{6x^3}{(x^4+1{)}^2}\mathrm{d}x$

Let $u=x^4+1$, then $du=4x^{3}dx$.

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{6x^3}{(x^4+1{)}^2}\mathrm{d}x& =\underset{t\to \mathrm{\infty }}{lim}{\int }_{0^4+1}^{t^4+1}\frac{3}{2u^2}\mathrm{d}u\\ & =\underset{t\to \mathrm{\infty }}{lim}\frac{3}{2}[-u^{-1}{]}_{0^4+1}^{t^4+1}\\ & =\underset{t\to \mathrm{\infty }}{lim}\frac{3}{2}(-(t^4+1{)}^{-1}+(0^4+1{)}^{-1})\\ & =\frac{3}{2}\end{array}$$

${\int }_{-\mathrm{\infty }}^2\sqrt{6-y}\mathrm{d}y$

Let $u=6-y$, then $du=-dy$.

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^2\sqrt{6-y}\mathrm{d}y& =\underset{t\to \mathrm{\infty }}{lim}{\int }_{6+t}^4\sqrt{u}\cdot -\mathrm{d}u\\ & =\underset{t\to \mathrm{\infty }}{lim}{[-\frac{2}{3}{u}^{\frac{3}{2}}]}_{6+t}^4\\ & =\underset{t\to \mathrm{\infty }}{lim}(-\frac{2}{3}{4}^{\frac{3}{2}}+\frac{2}{3}(6+t{)}^{\frac{3}{2}})\\ & =\mathrm{\infty }\end{array}$$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^3\mathrm{d}x$

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^3\mathrm{d}x& ={\int }_{-\mathrm{\infty }}^0{x}^3\mathrm{d}x+{\int }_0^{\mathrm{\infty }}{x}^3\mathrm{d}x\\ & =\underset{t\to -\mathrm{\infty }}{lim}{\int }_t^0{x}^3\mathrm{d}x+\underset{s\to \mathrm{\infty }}{lim}{\int }_0^s{x}^3\mathrm{d}x\\ & ={\underset{t\to -\mathrm{\infty }}{lim}\frac{1}{4}{x}^4|}_t^0+{\underset{s\to \mathrm{\infty }}{lim}\frac{1}{4}{x}^4|}_0^s\\ & =\underset{t\to \mathrm{\infty }}{lim}(0-\frac{1}{4}{t}^4)+\underset{s\to \mathrm{\infty }}{lim}(\frac{1}{4}{s}^4-0)\\ & =\text{D.N.E.}\end{array}$$

Contributors

Ajitani Hifumi

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