A few questions before my midterm on Friday

[Jamin2112]

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.

Homework Equations

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The Attempt at a Solution

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[hunt_mat]

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

 

[losiu99]

1) The coolest way I've ever seen is in Rudin's "Principles":
We use Bohr-Mollerup theorem to show that
\int_0 ^1 t^{x-1}(1-t)^{y-1}\,dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}
Substitute t=\sin^2 \theta, plugging x=y=1/2 gives \Gamma(1/2)=\sqrt{\pi}, which means that
\int_0^{\infty} t^{-1/2}e^{-t}\, dt =\sqrt{\pi}
Substitute t=s^2:
2\int_0^{\infty}e^{-s^2}\,ds = \sqrt{\pi}

Hope you enjoy this as much as I did.

 

[hunt_mat]

But how do you show that \Gamma (1/2)=\sqrt{\pi}?

 

[lanedance]

i'm guessing if you follow through the integration you end up with pi?
<br /> \int_0 ^1 t^{-1/2}(1-t)^{-1/2}\,dt=\frac{\Gamma(1/2)^2}{\Gamma(1)}<br />

 

[hunt_mat]

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.

 

[losiu99]

Well, if you substitute t=\sin^2 \theta, you end up with
2\int_0^{\pi / 2}\sin^{2x-1}\theta\cos^{2y-1}\theta\,d\theta
Letting x=y=1/2 we get
2\int_0^{\pi/2}d\theta=\pi

 

[hunt_mat]

Okay, this is a new way I have learned today. Thanks.

 

[Jamin2112]

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.

 

[hunt_mat]

I suspect that it means that they are supposed to be continuous on the range of trhe integral. Which is what you have.

 

[Jamin2112]

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.

 

[hunt_mat]

True but the integral over (0,1) is the same as the integral over [0,1].