Improper Riemann Integrals
How to integrate on unbounded domains.
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Study Improper Riemann Integrals
So far, we have integrated bounded functions on compact intervals \([a,b]\). In this part we skip these two assumptions by extending the integral notion to functions that may have a pole and/or are defined on unbounded intervals.
Unbounded Interval
Definition 1 (Integration on unbounded intervals). Let \(f:[a,\infty)\to\mathbb{R}\) be a function with the property that for all \(b>a\) the restriction of \(f\) to \([a,b]\) belongs to \(\mathcal{R}([a,b])\). If \[\lim_{b\to\infty}\int_a^bf(x)dx\] exists, then we say that \[\int_a^\infty f(x)dx\] is convergent. Otherwise, we speak of divergence.
Example 2.
For integrating the function \(\exp(-x)\) on the interval \([0,\infty)\), we compute \[\begin{aligned} \int_0^\infty\exp(-x)dx=&\lim_{b\to\infty}\int_0^b\exp(-x)dx\\=&\lim_{b\to\infty}\left.-\exp(-x)\right|_{x=0}^{x=b}\\ =&1-\lim_{b\to\infty}\exp(-b)=1. \end{aligned}\]
For \(\alpha>0\), consider \[\int_1^\infty \frac1{x^\alpha}dx.\] We know that for \(b>1\) holds \[\int_1^b\frac1{x^\alpha}dx=\begin{cases}\left.\frac1{1-\alpha}\frac1{x^{\alpha-1}}\right|_{x=1}^{x=b}&:\alpha\neq1,\\\left.\log(x)\right|_{x=1}^{x=b}&:\alpha=1. \end{cases}\] Since \(\lim_{b\to\infty}\log(b)=\infty\) and \[\lim_{b\to\infty}\frac1{x^{\alpha-1}}=\begin{cases}0&:\alpha>1,\\\infty&:\alpha<1,\end{cases}\] we have \[\int_1^\infty \frac1{x^\alpha}dx=\begin{cases}\frac1{\alpha-1}&:\alpha>1,\\\infty&:\alpha\leq1.\end{cases}\]
It is straightforward to define the integral of a function defined on some interval unbounded from below by \[\int_{-\infty}^af(x)dx=\lim_{b\to-\infty}\int_{b}^af(x)dx.\] We now define the integral of functions defined on the whole real axis.
Definition 3. Let \(f:\mathbb{R}\to\mathbb{R}\) be a function with the property that for all \(a,b\in\mathbb{R}\) the restriction of \(f\) to \([a,b]\) belongs to \(\mathcal{R}([a,b])\). If there exists some \(c\in\mathbb{R}\) such that both integrals \[\int_c^\infty f(x)dx,\qquad \int_{-\infty}^c f(x)dx\] exist, then we say that \[\int_{-\infty}^\infty f(x)dx\] is convergent. Otherwise, we speak of divergence. In case of convergence we set \[\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^c f(x)dx+\int_c^\infty f(x)dx.\]
We remark without proof that the above definition is independent of \(c\).
Example 4.
Consider \[\begin{aligned} \int_{-\infty}^\infty\frac1{1+x^2}dx=& \int_{-\infty}^0\frac1{1+x^2}dx+\int_0^\infty \frac1{1+x^2}dx\\ =&\lim_{a\to-\infty}\int_{a}^0\frac1{1+x^2}dx+\lim_{b\to\infty}\int_{0}^b\frac1{1+x^2}dx\\ =&\lim_{a\to-\infty}\left.\arctan(x)\right|_{x=a}^{x=0}+\lim_{b\to\infty}\left.\arctan(x)\right|_{x=0}^{x=b}\\ =&-\lim_{a\to-\infty}\arctan(a)+\lim_{b\to\infty}\arctan(b)\\ =&\frac\pi2+\frac\pi2=\pi. \end{aligned}\]
The integral \[\int_{-\infty}^\infty xdx\] diverges since both integrals \[\begin{aligned} \int_{-\infty}^0 xdx&=\lim_{a\to-\infty}\int_{a}^0 xdx=\lim_{a\to-\infty}\left.\frac12x^2\right|_{x=a}^{x=0}=-\infty\\ \int_{0}^\infty xdx&=\lim_{b\to\infty}\int_{0}^b xdx=\lim_{b\to\infty}\left.\frac12x^2\right|_{x=0}^{x=b}=\infty \end{aligned}\] diverge. This example shows that the convergence of \(\int_{-\infty}^\infty f(x)dx\) is not equivalent to the existence of the limit \[\lim_{a\to\infty}\int_{-a}^af(x)dx.\] However, in case of convergence, the integral \(\int_{-\infty}^\infty f(x)dx\) coincides with the above limit.
The majorant criterion for series says that if the absolut values of the addends of a given series can be bounded from above by the addends of a convergent series, then (absolut) convergence of the given series can be concluded.
Conversely, the minorant criterion for series says that if the addends of a given series can be bounded from below by the addends of a series that diverges to \(+\infty\), then also the given series diverges to \(+\infty\).
Analogue criteria hold true for integrals. We skip the proofs since they are totally analogous to those of the majorant and minorant criteria.
Theorem 5. Let \(f,g:[a,\infty)\to\mathbb{R}\) such that for all \(b\in [a,\infty)\), the restrictions of \(f\) and \(g\) to \([a,b]\) are Riemann-integrable.
If \(|f(x)|\leq g(x)\) for all \(x\in [a,\infty)\) and \(\int_{a}^\infty g(x)dx\) converges, then also \(\int_{a}^\infty f(x)dx\) converges and it holds that \[\left|\int_{a}^\infty f(x)dx\right|\leq \int_{a}^\infty |f(x)|dx\leq \int_{a}^\infty g(x)dx.\]
If \(g(x)\leq f(x)\) for all \(x\in [a,\infty)\) and \(\int_{a}^\infty g(x)dx=+\infty\), then \(\int_{a}^\infty f(x)dx=+\infty\).
Example 6.
Consider \[\int_1^\infty\frac{x}{x^2+1}dx.\] We have \[\lim_{x\to\infty}x\cdot\frac{x}{x^2+1}=1.\] For large enough \(x\in\mathbb{R}\), we therefore have \[\left|x\cdot\frac{x}{x^2+1}\right|=\frac{x^2}{x^2+1}\geq\frac12\] and thus \[\frac{x}{x^2+1}\geq\frac1{2x}.\] Since \[\int_1^\infty\frac{1}{2x}dx\] is divergent, \[\int_1^\infty\frac{x}{x^2+1}dx\] is divergent, too.
Consider \[\int_1^\infty\frac{\sqrt{x}}{x^2+1}dx.\] We have \[\lim_{x\to\infty}x^{3/2}\cdot\frac{\sqrt{x}}{x^2+1}=1.\] For large enough \(x\in\mathbb{R}\), we therefore have \[\left|x^{3/2}\cdot\frac{\sqrt{x}}{x^2+1}\right|=\frac{x^2}{x^2+1}\leq2\] and thus \[\frac{\sqrt{x}}{x^2+1}\leq\frac2{x^{3/2}}.\] Since \[\int_1^\infty\frac2{x^{3/2}}dx\] is convergent, so \[\int_1^\infty\frac{\sqrt{x}}{x^2+1}dx\] is convergent, too.
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