A few questions before my midterm on Friday

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Homework Help Overview

The discussion revolves around several calculus concepts, including the evaluation of integrals, uniform convergence, and the application of Dirichlet's Test. Participants are exploring various methods to prove specific integral results and clarifying definitions related to convergence.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss different methods to prove that the integral of e^(-x^2) converges to √π/2, including polar coordinates and the use of the Gamma function. Questions arise regarding the continuity of functions involved in Dirichlet's Test and the implications of uniform convergence.

Discussion Status

Several approaches to proving the integral results have been shared, with some participants expressing curiosity about the continuity conditions in Dirichlet's Test. There is an ongoing exploration of the implications of these mathematical concepts, with no explicit consensus reached yet.

Contextual Notes

Participants are working within the constraints of homework guidelines, which may limit the depth of exploration into certain proofs or methods. There is a noted emphasis on understanding rather than simply providing solutions.

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Homework Statement



1) What is a cool way to prove that ∫e-x2dx = √π / 2 ?

2) What does it mean if ∫f(x,t)dt converges uniformly on the interval a ≤ x ≤b? Does it basically mean that it'll converge for all x in that interval? Explain this to me as if I were an 8-year-old girl.

3) My book uses a variation of Dirichlet's Test to show that ∫sin(x) / x is convergent (bounds: 0, ∞). But if we use ø(x) = 1/x, like the book does, then isn't ø'(x) = -1/x2 not continuous? The book doesn't say anything about this.

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The Attempt at a Solution



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1) The only way I know is the usual way, by computing the square of the integral, and converting to polar co-ordinates.
2) Hmm, what are you varying? shoud there be and f_n around somewhere?

3) Because 1/x^2 is continuous on the OPEN inteval (0,infinity), it is only discontinuous at x=0
 
1) The coolest way I've ever seen is in Rudin's "Principles":
We use Bohr-Mollerup theorem to show that
[tex]\int_0 ^1 t^{x-1}(1-t)^{y-1}\,dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}[/tex]
Substitute [tex]t=\sin^2 \theta[/tex], plugging [tex]x=y=1/2[/tex] gives [tex]\Gamma(1/2)=\sqrt{\pi}[/tex], which means that
[tex]\int_0^{\infty} t^{-1/2}e^{-t}\, dt =\sqrt{\pi}[/tex]
Substitute [tex]t=s^2[/tex]:
[tex]2\int_0^{\infty}e^{-s^2}\,ds = \sqrt{\pi}[/tex]

Hope you enjoy this as much as I did.
 
But how do you show that [tex]\Gamma (1/2)=\sqrt{\pi}[/tex]?
 
i'm guessing if you follow through the integration you end up with pi?
[tex] \int_0 ^1 t^{-1/2}(1-t)^{-1/2}\,dt=\frac{\Gamma(1/2)^2}{\Gamma(1)}[/tex]
 
The only way that I know of showing that the integral is pi, is the one that everyone gets taught by squaring the integral, turning it into a double integral, transforming to polar co-ordinates and then solving the integral.
 
Well, if you substitute [tex]t=\sin^2 \theta[/tex], you end up with
[tex]2\int_0^{\pi / 2}\sin^{2x-1}\theta\cos^{2y-1}\theta\,d\theta[/tex]
Letting [tex]x=y=1/2[/tex] we get
[tex]2\int_0^{\pi/2}d\theta=\pi[/tex]
 
Okay, this is a new way I have learned today. Thanks.
 
hunt_mat said:
3) Because 1/x^2 is continuous on the OPEN inteval (0,infinity), it is only discontinuous at x=0


But in the book's description of the Dirichlet's Test, if you have ∫f(x)ø(x)dx, then one of the conditions is that "ø'(x) is continuous." That's all it says.
 
  • #10
I suspect that it means that they are supposed to be continuous on the range of trhe integral. Which is what you have.
 
  • #11
hunt_mat said:
I suspect that it means that they are supposed to be continuous on the range of trhe integral. Which is what you have.

The integral goes from 0 to infiniti, and ø'(t) is not continuous at x = 0.
 
  • #12
True but the integral over (0,1) is the same as the integral over [0,1].
 

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