improper integral

Improper integral

An improper integral is an extension of the notion of a definte integral to cases that violate the usual assumptions of integrals. For example, in the context of Riemann integrals, improper integrals typically involve unboundedness over the interval of integration and/or the integrand. It may also involve bounded but open sets or bounded but discontinuous functions.

Integrals with infinite limits of integration

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  • If \[f(x)\] is continuous on \[\left[ a,\infty \right)\], then \[\int_{a}^{\infty}f(x)\,dx=\lim_{b\to\infty}\int_{a}^{b}f(x)\,dx\].
  • If \[f(x)\] is continuous on \[\left( -\infty,b \right]\], then \[\int_{-\infty}^{b}f(x)\,dx=\lim_{a\to\infty}\int_{-a}^{b}f(x)\,dx\].
  • If \[f(x)\] is continuous on \[\left( -\infty,\infty \right)\], then \[\int_{-\infty}^{\infty}f(x)\,dx=\int_{-\infty}^{c}f(x)\,dx+\int_{c}^{\infty}f(x)\,dx\], where \[c\] is any real number.

As an example, evaluate \[\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}\,dx\].

Let \[c=0\], then \[\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}\,dx=\int_{-\infty}^{0}\frac{1}{1+x^{2}}\,dx+\int_{0}^{\infty}\frac{1}{1+x^{2}}\,dx\]. Therefore,

\begin{align*} \int_{-\infty}^{0}\frac{1}{1+x^{2}}\,dx&=\lim_{a\to-\infty}\int_{a}^{0}\frac{1}{1+x^{2}}\,dx\\ &=\lim_{a\to-\infty}\biggl[ \tan^{-1}(x) \biggr]_{a}^{0}\\ &=\lim_{a\to-\infty}\left[ \tan^{-1}(0)-\tan^{-1}(a) \right]\\ &=\lim_{a\to-\infty}-\tan^{-1}(a)\\ &=\frac{\pi}{2}\\ \end{align*}

Similarly,

\begin{align*} \int_{0}^{\infty}\frac{1}{1+x^{2}}\,dx&=\lim_{b\to\infty}\int_{0}^{b}\frac{1}{1+x^{2}}\,dx\\ &=\lim_{b\to\infty}\biggl[ \tan^{-1}(x) \biggr]_{0}^{b}\\ &=\lim_{b\to\infty}\left[ \tan^{-1}(b)-\tan^{-1}(0) \right]\\ &=\lim_{b\to\infty}\tan^{-1}(b)\\ &=\frac{\pi}{2}\\ \end{align*}

So, \[\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}=\frac{\pi}{2}+\frac{\pi}{2}=\pi\].

Integrands with vertical asymptotes

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  • If \[f(x)\] is continuous on \[\left( a,b \right]\] and discontinuous at \[a\], then \[\int_{a}^{b}f(x)\,dx=\lim_{c\to a^{+}}\int_{c}^{b}f(x)\,dx\].
  • If \[f(x)\] is continuous on \[\left[ a,b \right)\] and discontinuous at \[b\], then \[\int_{a}^{b}f(x)\,dx=\lim_{c\to b^{-}}\int_{a}^{c}f(x)\,dx\].
  • If \[f(x)\] is continuous at \[c\], where \[a<c<b\], but is still continuous on \[\left[ a,c \right)\cup \left( c,b \right]\], then \[\int_{a}^{b}f(x)\,dx=\int_{a}^{c}f(x)\,dx+\int_{c}^{b}f(x)\,dx\].

As an example, evaluate \[\int_{0}^{1}\frac{1}{1-x}\,dx\].

\begin{align*} \int_{0}^{1}\frac{1}{1-x}\,dx&=\lim_{b\to1^{-}}\int_{0}^{b}\frac{1}{1-x}\,dx\\ &=\lim_{b\to1^{-}}-\int_{0}^{b}\frac{1}{x-1}\,dx\\ &=\lim_{b\to1^{-}}\biggl[ -\ln \left| x-1 \right| \biggr]_{0}^{b}\\ &=\lim_{b\to1^{-}}\biggl[ -\ln \left( x-1 \right) \biggr]_{0}^{b}\\ &=\lim_{b\to1^{-}}\left[ -\ln(1-b) \right]\\ &=-(-\infty)\\ &=\infty \end{align*}

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