Use the comparison test to show the integrals are convergent and divergent respectively.?
Use the comparison test to show the integral Int[1 to infinity] (sin^2 x)/x^3 dx is convergent and Int[2 to infinity] 1/(ln x) dx is divergent. I'm not sure about the comparison test used to determine the convergence or divergence of an improper integral, I have only heard about that test for infinite series only, but not integrals. Maybe I could replace the integral operator by the summation notation from x = 1 to oo and let x be discrete. This is the same as doing the integral test for the infinite series. This question is from this page http://www.rit.edu/~bpbsma/242/improper.pdf
Improper Integrals: Which of these integrals converge?
Which of the following four integrals: x/(1+x^4) from 0 to inf x^2/(1+x^4) from 0 to inf x^3/(1+x^4) from 0 to inf x^4/(1+x^4) from 0 to inf Which of these integrals converge and what is logic behind their convergence? (Hint: compare to "pure powers of x.) Compute the exact value of at least one of the convergent integrals.
How can I use the comparison test to determine the convergence of these integrals?
Note that [math]\log(x) < x[/math] for all [math]x \geq 3[/math]. Therefore, we have for (a):[math]\displaystyle \int_{3}^{n} \frac{dx}{\log(x)} > \int_{3}^{n} \frac{dx}{x} = \log(n) - \log(3)[/math]As the rightmost term grows without bound as [math]n\to \infty[/math], our original integral must be divergent.For (b), observe that the integrand has a pole at [math]x=1[/math] and therefore the integral doesn't exist. However, the integral has a computable Cauchy principal value:[math]\displaystyle \lim_{\epsilon \to 0} \int_{0}^{1 - \epsilon} \frac{dx}{\log(x)} + \int_{1 + \epsilon}^{3} \frac{dx}{\log(x)} = Li(3) \approx 2.163[/math](You should prove that this limit exists. Hint: let [math]F(x)[/math] be an antiderivative of the integrand. Try to put bounds on the expression [math]F(1 + \epsilon) - F(1 - \epsilon)[/math] using the mean value theorem.)For (c), the integral is once again undefined due to the pole at [math]x=1[/math], but this time we do not have a Cauchy principal value either as the conditions for its existence are no longer satisfied. (You may check this yourself.)For (d), note that [math]f(x) = x^4 e^{-x^2}[/math] is decreasing for [math]x \geq \sqrt{2}[/math] and we have that [math]f(\sqrt{2}) = 4 e^{-2} < 1[/math], which implies that [math]f(x) < 1[/math] for all [math]x[/math]. Therefore, we have the following estimates:[math]\displaystyle \int_{-\infty}^{\infty} x^2 e^{-x^2}\, dx = \int_{-1}^{1} x^2 e^{-x^2}\, dx + 2\int_{1}^{\infty} x^2 e^{-x^2}\, dx < K + 2\int_{1}^{\infty} x^{-2}\, dx = K + 2[/math]where [math]K[/math] is a finite constant. As the integral is bounded above by [math]K+2[/math] and below by [math]0[/math], it converges.
How do you use the Comparison Theorem to show sec^2(X)/(x*sqrt(x)) diverges?
This integral diverges. We can show this by the Comparison Test. Note that sec^2(x) ≥ 1 for all x. So, sec^2(x) / (x√x) ≥ 1/(x√x) = x^(-3/2) for all x > 0. Since ∫(x = 0 to 1) x^(-3/2) dx = lim(t→0+) ∫(x = t to 1) x^(-3/2) dx = lim(t→0+) -2x^(-1/2) {for x = t to 1} = lim(t→0+) 2(1/t^(1/2) - 1) = ∞ (and thus divergent), we conclude that ∫(x = 0 to 1) sec^2(x) dx/x^(3/2) is also divergent. I hope this helps!
Find all values p for which the following integral converges:
You can do this by recalling that [math]\ln' x = \frac{1}{x}[/math]Let [math]u=\ln x[/math]. Then [math]du=\frac{dx}{x}[/math] and the integral becomes [math]\int_{\ln e}^{\infty} \frac{du}{u^p}[/math]For [math]p>1[/math], [math]\int_{\ln e}^{\infty} \frac{du}{u^p}=\frac{1}{(1-p)u^{p-1}}\big{|}^{\infty}_1[/math], which converges. For [math]p\leq1[/math], the integral diverges.
Convergence and Divergence using Comparison Test?
Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test). For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) 1. For all n>1, 1/(nln(n))<2n, and the series 2∑1/n diverges, so by the Comparison Test, the series ∑1/(nln(n)) diverges 2. For all n>2, n/(n^3−8)<2n^2, and the series 2∑1/n^2 converges, so by the Comparison Test, the series ∑n/(n^3−8) converges. 3. For all n>1, arctan(n)/(n^3)<π/(2n^3), and the series π^2∑1/n^3 converges, so by the Comparison Test, the series ∑arctan(n)/(n^3) converges. 4. For all n>1, sin^2(n)/n^2<1/n^2, and the series ∑1/n^2 converges, so by the Comparison Test, the series ∑sin^2(n)/n^2 converges. 5. For all n>1, n^5−n^3<1/n^2, and the series ∑1/n^2 converges, so by the Comparison Test, the series ∑n^5−n^3 converges. 6. For all n>2, 1/(n^2−4)<1/n^2, and the series ∑1/n^2 converges, so by the Comparison Test, the series ∑1/(n^2−4) converges. I got C, C, I, C, C, I. The computer says its incorrect but I'm not sure which one is wrong.
Check if the following integrals converge or diverge, and find their values, if possible?
a) Since the integrand has an infinite discontinuity at x = 0, ∫(x = -π/2 to π/2) csc x dx = ∫(x = -π/2 to 0) csc x dx + ∫(x = 0 to π/2) csc x dx. So, the integral in question converges iff each term converges. However, ∫(x = 0 to π/2) csc x dx = lim(t→0+) ∫(x = t to π/2) csc x dx = lim(t→0+) -ln |csc x + cot x| {for x = t to π/2} = lim(t→0+) ln(csc t + cot t) = lim(t→0+) ln((1 + cos x)/sin x) = ∞. Hence, the integral in question is divergent. (Note: If you are assuming principal value, then symmetry dictates that the integral equals 0. My hunch is that this is not the case.) ----------------- b) Use the Weierstrauss substitution t = tan(x/2). So, sin x = 2t/(1 + t^2), cos x = (1 - t^2)/(1 + t^2), dx = 2 dt/(1 + t^2). Link: http://en.wikipedia.org/wiki/Weierstrass... Therefore, ∫(x = 0 to π/2) cos x dx / (1 + sin x - cos x) = ∫(t = 0 to 1) [(1 - t^2)/(1 + t^2)] * [2 dt/(1 + t^2)] / [1 + 2t/(1 + t^2) - (1 - t^2)/(1 + t^2)] = ∫(t = 0 to 1) 2(1 - t^2) dt / {(1 + t^2) [(1 + t^2) + 2t - (1 - t^2)]} = ∫(t = 0 to 1) (1 - t^2) dt / [(1 + t^2) (t^2 + t)] = lim(r→0+) ∫(t = r to 1) [1/t - t/(t^2 + 1) - 1/(t^2 + 1)] dt, by partial fractions = lim(r→0+) [ln t - (1/2) ln(t^2 + 1) - arctan t] {for t = r to 1} = lim(r→0+) [ln (t/√(t^2 + 1)) - arctan t] {for t = r to 1} = lim(r→0+) [(ln(1/√2) - π/4) - (ln (r/√(r^2 + 1)) - arctan r)] = ∞, since lim(r→0+) r/√(r^2 + 1) = 0. ----------------- c) ∫(x = 0 to ∞) |sin x| dx/x^2 is improper at both 0 and ∞. So, we split up the integral, to deal with one improper point at a time: ∫(x = 0 to 1) |sin x| dx/x^2 + ∫(x = 1 to ∞) |sin x| dx/x^2 = ∫(x = 0 to 1) sin x dx/x^2 + ∫(x = 1 to ∞) |sin x| dx/x^2. Although the second term converges (using the comparison |sin x|/x^2 ≤ 1/x^2), the first term diverges. Hence the integral in question diverges. To show that ∫(x = 0 to 1) sin x dx/x^2 diverges: ∫(x = 0 to 1) sin x dx/x^2 = ∫(∞ to 1) sin(1/w) * (-dw/w^2) / (1/w^2), letting w = 1/x = ∫(w = 1 to ∞) sin(1/w) dw. This diverges via Limit Comparison Test: lim(w→∞) sin(1/w) / (1/w) = lim(t→0+) sin(t)/t, letting t = 1/w = 1, but ∫(w = 1 to ∞) (1/w) dw = ln w {for w = 1 to ∞} = ∞ (divergent). --------------------- I hope this helps!
How can I determine whether an improper integral diverges without solving the integral?
You can use what is known as the Comparison Theorem for Integrals.Basically, if you’re taking the integral of some function f(x) which is either too difficult to solve or for which you are only asked about the convergence/divergence of it, you find some similar (but simpler) funtion g(x) to compare it with.If you know that g(x) is less than or equal to f(x) on the interval of integration, and you find that g(x) diverges, then it must be the case that f(x) also diverges since its value is larger than that of g(x) on the interval.Now if g(x) is greater than or equal to f(x) on the interval of integration, and you find that g(x) converges, then it must also be the case that f(x) converges since the convergence of the larger function g(x) implies that any similar function of lesser value will behave in the same manner on the given interval.