Improper integral sinx/sqrt(x) from 1 to infinity?
The curve sin(x)/sqrt(x) has alternating 'humps' above and below the x-axis, each with a smaller area than the preceding hump. These areas converge to 0 as x -> infinity. The integral can thus be written as an alternating series with terms that go to 0 as x -> infinity, and thus by the alternating series test the integral converges. This was probably too informal, so I'll try and come up with something more rigorous shortly. One more quick note for now: if we let u = sqrt(x) then the integral becomes integral(u=1 to infinity)(2*sin(u^2) du), which is a Fresnel integral and has a known value of 0.632778 to 6 decimal places. Edit: Use integration by parts. Let u = x^(-1/2) and dv = sin(x). Then du = (-1/2)*x^(-3/2) dx and v = -cos(x), and so integral((sin(x)/sqrt(x)) dx) = -x^(-1/2)*cos(x) - (1/2)*integral(cos(x)*x^(-3/2) dx). When this is evaluated from x = 1 to x = infinity we end up with lim(x->infinity)(-x^(-1/2)*cos(x)) + cos(1) - (1/2)*integral(cos(x)*x^(-3/2) dx) (from x = 1 to x = infinity). Now the first limit is 0, and the remaining integral converges by comparison to the integral(x=1 to infinity)(x^(-3/2) dx) since -1 <= cos(x) <= 1 for all x. (Note that integral(x=1 to infinity)(x^(-3/2) dx) = 2.)
How do I solve integral from 0 to infinity cos(x^2)?
Are you aware of the Gaussian integral? From the wiki,[math]\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}[/math]The wiki provides multiple proofs for the same, so I won't replicate it here. We will be using this to get the value of the desired definite integral.Claim 1: [math]\int_{0}^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}[/math]Proof: Since [math]e^{-x^2}[/math] is an even function.Claim 2: [math]\int_{0}^{\infty}e^{ix^2}dx = \frac{e^{i\pi / 4}}{2}\sqrt{\pi}[/math]Proof: Let [math]I = \int_{0}^{\infty}e^{ix^2}dx[/math]. Substituting [math]x = e^{i\pi / 4}y[/math], we get,[math]I = \int_{0}^{\infty}e^{i\pi / 4}e^{-y^2}dy = \frac{e^{i\pi / 4}}{2}\sqrt{\pi}[/math]Note that the proof of Claim 2 is not entirely rigorous. Ideally, we have to argue that the complex integral is indeed the same as integrating over the positive real axis. I am somewhat glossing over it at the moment, but the argument is standard, so if you have experience in complex integration, you should be able to fill it in.Claim 3: [math]\int_{0}^{\infty}\cos x^2 dx = \sqrt{\frac{\pi}{8}}[/math]Proof: From Claim 2,[math]\int_{0}^{\infty}e^{ix^2}dx = \frac{e^{i\pi / 4}}{2}\sqrt{\pi}[/math][math]\int_{0}^{\infty}\cos x^2 dx + i \int_{0}^{\infty}\sin x^2 dx = \frac{\sqrt{\pi}}{2}\left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right)[/math]Comparing the real parts, gives us the required answer.
Determine if improper integral, from 0 to 1, of 4ln3x / sqrtx dx is convergent or divergent, and if...?
∫4ln(3x) dx /sqrt(x) multiply and divide by sqrt(x) 4∫[sqrt(x) ln(3x)/x]dx let ln(3x) = y, then e^y = 3x : x = (1/3)e^y : sqrt(x) = 1/sqrt(3)e^(y/2) (1/x)dx = dy now the integral becomes 4∫(1/sqrt3)e^(y/2) y dy =>4/sqrt3∫y e^(y/2) dy integrating by parts u = (4/sqrt3)y : du = (4/sqrt3)dy dv = e^(y/2) : v = 2e^(y/2) 4/sqrt3∫u e^u du = uv - ∫v du =>(8/sqrt3)y e^(y/2) -8/sqrt3 ∫e^(y/2) dy =>(8/sqrt3)y e^(y/2) - (16/sqrt3) e^(y/2) =>(8/sqrt3)e^(y/2)[y - 2] substitute back y = ln(3x) and e^(y/2) = sqrt(3x) (8/sqrt3)sqrt(3x)[ln(3x) -2] =>8sqrt(x)[ln(3x) - 2] from [0 to 1] 8sqrt(1)[ln(3) - 2) 8[ln(3) - 2]
The Improper Integral from 0 to 1 of 3xlnxdx ANY HELP PLEASE?!?
1 ............. .........1 ∫3x lnx dx = lim... ∫ 3x lnx dx 0 .............. t->0 ..t integral by parts: let u = lnx ; dv = 3x du 1/x dx ; v = 3x²/2 3x²/2 lnx - ∫3x²/2 (1/x) dx 3x²/2 lnx - ∫3x/2 3x²/2 lnx - 3x²/4 Use the FTC, and you'll get: -3/4 - 3t²/2 lnt + 3t²/4 take the limit as t > 0 lim -3/4 - lim 3t²/2 lnt + lim 3t²/4 t->0 .......t->0 ............. t->0 to find the limit in the middle term, use L'hospital rule: 3t²/2 lnt = (3lnt) / (2/t²) d/dt (3lnt) / d/dt (2/t²) = (3/t) / (-2/t^3) = 3/t * (-t^3 / 2) = (-3/2)t² as t approaches 0, (-3/2)t² also appreaches 0 hence, lim -3/4 - lim 3t²/2 lnt + lim 3t²/4 t->0 .......t->0 ............. t->0 = -3/4 - 0 + 0 = -3/4 hope it helps! =]
Using polar coordinates, evaluate the improper integral...?
(A) Thinking of R² as the limiting region of a circle centered at the origin with radius R and letting R go to infinity: ∫∫(R²) e^[-4(x² + y²)] dx dy = ∫(θ = 0 to 2π) ∫(r = 0 to ∞) e^(-4r²) (r dr dθ) = 2π ∫(r = 0 to ∞) r e^(-4r²) dr = 2π * (-1/8) e^(-4r²) {for r = 0 to ∞} = π/4 * (0 - (-1)) = π/4. B) Note that ∫∫(R²) e^[-4(x² + y²)] dx dy = {∫(-∞ to ∞) e^(-4x²) dx} * {∫(-∞ to ∞) e^(-4y²) dy} = {∫(-∞ to ∞) e^(-4x²) dx}². Thus, {∫(-∞ to ∞) e^(-4x²) dx}² = π/4, by (A). Since the integrand is positive, taking positive square roots of both sides yields ∫(-∞ to ∞) e^(-4x²) dx = √π / 2. I hope this helps!
Determine whether the improper integral is convergent or divergent without solving the integral (3 to infinity) (tan^(-1) x)/x^4 dx.?
Use the Comparison Test. First of all, the integrand is continuous and positive for all x > 0 (and thus for the region of integration). Moreover, for all x > 3, we have arctan x < π/2 ==> arctan(x)/x^4 < (π/2)/x^4. Since ∫(x = 3 to ∞) (π/2) dx/x^4 = (-π/6) x^(-3) {for x = 3 to ∞} = π/162 (and thus convergent), we conclude that the integral in question also converges by the Comparison Test. I hope this helps!