We show that a variety oftrigonometric sums have unexpected closed forms by relatingthem to cognate integrals. The following chapter introduces us to a number of different problems whose solution is provided by integration. % \(e^{-x} \ll x^2\text{,}\) so that the denominator \(e^{-x}+x^2\approx x^2\text{,}\) and, \(|\sin x|\le 1 \ll x\text{,}\) so that the numerator \(x+\sin x\approx x\text{,}\) and, the integrand \(\frac{x+\sin x}{e^{-x}+x^2} \approx \frac{x}{x^2} =\frac{1}{x}\text{.}\). a To this point we have only considered nicely behaved integrals \(\int_a^b f(x)\, d{x}\text{. Both of these are examples of integrals that are called Improper Integrals. {\textstyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}} This content iscopyrighted by a Creative CommonsAttribution - Noncommercial (BY-NC) License. Being able to compare "unknown" integrals to "known" integrals is very useful in determining convergence. x The second one can be addressed by calculus techniques, but also in some cases by contour integration, Fourier transforms and other more advanced methods. Let \(f\) and \(g\) be continuous on \([a,\infty)\) where \(0\leq f(x)\leq g(x)\) for all \(x\) in \([a,\infty)\). An improper integral is said to converge if its corresponding limit exists; otherwise, it diverges. Notice that we are using \(A \ll B\) to mean that \(A\) is much much smaller than \(B\). Let's eschew using limits for a moment and proceed without recognizing the improper nature of the integral. The purpose of using improper integrals is that one is often able to compute values for improper integrals, even when the function is not integrable in the conventional sense (as a Riemann integral, for instance) because of a singularity in the function as an integrand or because one of the bounds of integration is infinite. \begin{align*} \int_0^1\frac{\, d{x}}{x} &=\lim_{t\rightarrow 0+}\int_t^1\frac{\, d{x}}{x} =\lim_{t\rightarrow 0+}\Big[\log x\Big]_t^1 =\lim_{t\rightarrow 0+}\log\frac{1}{t} =+\infty\\ \int_{-1}^0\frac{\, d{x}}{x} &=\lim_{T\rightarrow 0-}\int_{-1}^T\frac{\, d{x}}{x} =\lim_{T\rightarrow 0-}\Big[\log|x|\Big]_{-1}^T =\lim_{T\rightarrow 0-}\log|T|\ =-\infty \end{align*}. f The improper integral can also be defined for functions of several variables. This, too, has a finite limit as s goes to zero, namely /2. This will, in turn, allow us to deal with integrals whose integrand is unbounded somewhere inside the domain of integration. f }\) The given integral is improper. {\displaystyle 1/{x^{2}}} \[\int_{{\,a}}^{{\,\,b}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{t \to {b^ - }} \int_{{\,a}}^{{\,t}}{{f\left( x \right)\,dx}}\], If \(f\left( x \right)\) is continuous on the interval \(\left( {a,b} \right]\) and not continuous at \(x = a\) then, An improper integral of the first kind. PDF Math 104: Improper Integrals (With Solutions) - University of Pennsylvania For what values of \(p\) does \(\int_e^\infty\frac{\, d{x}}{x(\log x)^p}\) converge? }\) Recall that the first step in analyzing any improper integral is to write it as a sum of integrals each of has only a single source of impropriety either a domain of integration that extends to \(+\infty\text{,}\) or a domain of integration that extends to \(-\infty\text{,}\) or an integrand which is singular at one end of the domain of integration. Does the integral \(\displaystyle\int_0^\infty\frac{x+1}{x^{1/3}(x^2+x+1)}\,\, d{x}\) converge or diverge? Replacing 1/3 by an arbitrary positive value s (with s < 1) is equally safe, giving /2 2 arctan(s). \begin{align*} f(x) &= \frac{x+\sin x}{e^{-x}+x^2} & g(x) &= \frac{1}{x} \end{align*}, \begin{align*} \lim_{x\rightarrow\infty}\frac{f(x)}{g(x)} &=\lim_{x\rightarrow\infty} \frac{x+\sin x}{e^{-x}+x^2}\div\frac{1}{x}\\ &=\lim_{x\rightarrow\infty} \frac{(1+\sin x/x)x}{(e^{-x}/x^2+1)x^2}\times x\\ &=\lim_{x\rightarrow\infty} \frac{1+\sin x/x}{e^{-x}/x^2+1}\\ &=1 \end{align*}. {\displaystyle \mathbb {R} ^{n}} }\), \begin{align*} \int_t^1 \frac{1}{x}\, d{x} &= \log|x| \bigg|_t^1 = -\log|t| \end{align*}, \begin{align*} \int_0^1 \frac{1}{x}\, d{x} &= \lim_{t=0^+}\int_t^1 \frac{1}{x}\, d{x} = \lim_{t=0^+} -\log|t| = +\infty \end{align*}. The domain of the integral \(\int_1^\infty\frac{\, d{x}}{x^p}\) extends to \(+\infty\) and the integrand \(\frac{1}{x^p}\) is continuous and bounded on the whole domain. This is described in the following theorem. Can anyone explain this? Answer: 42) 24 6 dt tt2 36. Thus, for instance, an improper integral of the form, can be defined by taking two separate limits; to wit. The prior analysis can be taken further, assuming only that G(x) = 0 for x / (,) for some > 0. However, any finite upper bound, say t (with t > 1), gives a well-defined result, 2 arctan(t) /2. Somehow the dashed line forms a dividing line between convergence and divergence. }\) A good way to start is to think about the size of each term when \(x\) becomes big. is as n approaches infinity. However, there are integrals which are (C,) summable for >0 which fail to converge as improper integrals (in the sense of Riemann or Lebesgue). + This is an innocent enough looking integral. This limit doesnt exist and so the integral is divergent. {\displaystyle f_{+}=\max\{f,0\}} Using L'Hpital's Rule seems appropriate, but in this situation, it does not lead to useful results. Does the integral \(\displaystyle\int_{-\infty}^\infty \sin x \, d{x}\) converge or diverge? Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. To do so, we want to apply part (a) of Theorem 1.12.17 with \(f(x)= \frac{\sqrt{x}}{x^2+x}\) and \(g(x)\) being \(\frac{1}{x^{3/2}}\text{,}\) or possibly some constant times \(\frac{1}{x^{3/2}}\text{. At the lower bound of the integration domain, as x goes to 0 the function goes to , and the upper bound is itself , though the function goes to 0. We have: \[\begin{align} \lim_{b\to\infty}\frac{\ln b}b &\stackrel{\ \text{ by LHR } \ }{=} \lim_{b\to\infty} \frac{1/b}{1} \\ &= 0.\end{align}\], \[\int_1^\infty\frac{\ln x}{x^2}\ dx = 1.\]. You can play around with different functions and see which ones converge or diverge at what rates. The limit exists and is finite and so the integral converges and the integrals value is \(2\sqrt 3 \). {\displaystyle [-a,a]^{n}} Let's now formalize up the method for dealing with infinite intervals. But If \(|f(x)|\le g(x)\) for all \(x\ge a\) and if \(\int_a^\infty g(x)\, d{x}\) converges then \(\int_a^\infty f(x)\, d{x}\) also converges. If we use this fact as a guide it looks like integrands that go to zero faster than \(\frac{1}{x}\) goes to zero will probably converge. }\) Suppose \(\displaystyle\int_0^\infty f(x) \, d{x}\) converges, and \(\displaystyle\int_0^\infty g(x) \, d{x}\) diverges. yields an indeterminate form, Direct link to lzmartinico's post What is a good definition, Posted 8 years ago. Legal. Figure \(\PageIndex{12}\): Graphing \(f(x)=\frac{1}{\sqrt{x^2+2x+5}}\) and \(f(x)=\frac1x\) in Example \(\PageIndex{6}\). \begin{align*} \int_e^\infty\frac{\, d{x}}{x(\log x)^p} &=\lim_{R\rightarrow \infty} \int_e^R\frac{\, d{x}}{x(\log x)^p} \qquad\qquad\qquad \text{use substitution}\\ &=\lim_{R\rightarrow \infty} \int_1^{\log R}\frac{\, d{u}}{u^p} \qquad\qquad\text{with }u=\log x,\, d{u}=\frac{\, d{x}}{x}\\ &=\lim_{R\rightarrow\infty} \begin{cases} \frac{1}{1-p}\Big[(\log R)^{1-p}-1\Big] & \text{if } p\ne 1\\ \log(\log R) & \text{if } p=1 \end{cases}\\ &=\begin{cases} \text{divergent} & \text {if } p\le 1\\ \frac{1}{p-1} & \text{if } p \gt 1 \end{cases} \end{align*}, The gamma function \(\Gamma(x)\) is defined by the improper integral, \[ \Gamma(t) = \int_0^\infty x^{t-1}e^{-x}\, d{x} \nonumber \], We shall now compute \(\Gamma(n)\) for all natural numbers \(n\text{. \[\int_{{\,a}}^{{\,\,b}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{t \to {a^ + }} \int_{{\,t}}^{{\,b}}{{f\left( x \right)\,dx}}\], If \(f\left( x \right)\) is not continuous at \(x = c\) where \(a < c < b\) and \( \displaystyle \int_{{\,a}}^{{\,c}}{{f\left( x \right)\,dx}}\) and \( \displaystyle \int_{{\,c}}^{{\,\,b}}{{f\left( x \right)\,dx}}\) are both convergent then, When dealing with improper integrals we need to handle one "problem point" at a time. , since the double limit is infinite and the two-integral method. n If \(\int_a^\infty g(x)\, d{x}\) converges, then the area of, If \(\int_a^\infty g(x)\, d{x}\) diverges, then the area of, So we want to find another integral that we can compute and that we can compare to \(\int_1^\infty e^{-x^2}\, d{x}\text{. \[\begin{align} \int_{-\infty}^\infty \frac1{1+x^2}\ dx &= \lim_{a\to-\infty} \int_a^0\frac{1}{1+x^2}\ dx + \lim_{b\to\infty} \int_0^b\frac{1}{1+x^2}\ dx \\ &= \lim_{a\to-\infty} \tan^{-1}x\Big|_a^0 + \lim_{b\to\infty} \tan^{-1}x\Big|_0^b\\ &= \lim_{a\to-\infty} \left(\tan^{-1}0-\tan^{-1}a\right) + \lim_{b\to\infty} \left(\tan^{-1}b-\tan^{-1}0\right)\\ &= \left(0-\frac{-\pi}2\right) + \left(\frac{\pi}2-0\right).\end{align}\] Each limit exists, hence the original integral converges and has value:\[= \pi.\] A graph of the area defined by this integral is given in Figure \(\PageIndex{5}\). f f Figure \(\PageIndex{9}\) graphs \(y=1/x\) with a dashed line, along with graphs of \(y=1/x^p\), \(p<1\), and \(y=1/x^q\), \(q>1\). There is more than one theory of integration. the antiderivative of 1 over x squared or x closer and closer to 0. Here is a theorem which starts to make it more precise. We have this area that {\textstyle 1/{\sqrt {x}}} Figure \(\PageIndex{2}\): A graph of \(f(x) = \frac{1}{x^2}\) in Example \(\PageIndex{1}\). approaches infinity of the integral from 1 to When does this limit converge -- i.e., when is this limit not \(\infty\)? , so, with Well, by definition Consequently, the integral of \(f(x)\) converges if and only if the integral of \(g(x)\) converges, by Theorems 1.12.17 and 1.12.20.
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