Show that if equivalent to

[Jamin2112]

Show that ... if equivalent to ...

Homework Statement

Show that

∫e^-t2sin(2xt)dt (bounds: 0,∞)

= e^-x2∫e^u2du (bounds: x, 0)

Homework Equations

THEOREM X. Let the integral F(x) = ∫f(t,x)dt (bounds: c, ∞) be convergent when a ≤ x ≤ b. Let the partial derivative ∂f/∂x be continuous in the two variables t, x when c ≤ t and a ≤ x ≤ b, and let the integral ∫∂f/∂x dt be uniformly convergent on [a, b]. Then F(x) has a derivative given by F'(x) = ∫ ∂f(t, x)/∂x dt.

The Attempt at a Solution

I think I need to differentiate both sides enough times until I've "shown" that the equality is true. So far, no success.

I'd let G(x) = ∫e^u2 du (bounds: 0, x). Then, using the fundamental theorem of calculus,

(e^-x2G(x))' = e^-x2 * e^x2 + e^u2 du * e^-x2.

Some pattern will emerge when I keep on differentiating.

As for ∫e^-t2sin(2xt)dt (bounds: 0,∞), I'll have 2x accumulating over and over again, and then an alternation between sin(2xt) and cos(2xt), etc...


Am I doing this right? I'm not finding any major cancellations.

 

[tiny-tim]

Hi Jamin2112!

∫e^-t2sin(2xt)dt (bounds: 0,∞)

Wouldn't it be easier to use Euler's formula (cos + isin), and then complete the square ?

 

[Jamin2112]

Hi Jamin2112!


Wouldn't it be easier to use Euler's formula (cos + isin), and then complete the square ?

First of all, should I be differentiating both sides of the original equation?

 

[tiny-tim]

Hi Jamin2112!

First of all, should I be differentiating both sides of the original equation?

Yes.

Differentiate the second one first (so that you know what you're aiming at! ), to get an expression for dF/dx as a function of F,

and then use theorem X on the first one, followed by integration by parts, to make it look like the first result.

What do you get? ​

 

[Jamin2112]

Hi Jamin2112!


Yes.

Differentiate the second one first (so that you know what you're aiming at! ), to get an expression for dF/dx as a function of F,

and then use theorem X on the first one, followed by integration by parts, to make it look like the first result.

What do you get? ​

Differentiating the second one yields

e^-x2 * e^-x2 + ∫e^u2du * -2x²e^-x2 = 1 - 2xe^-x2∫e^u2du.

So if G(x) = e^-x2∫e^u2du, I have

G' = 1 - 2x * G
G' / G = 1 - 2x
d/dx[ln(G)] = 1 - 2x
ln(G) = x - x² + C
G = e^{x - x2 + C}

Hmmmmm ... I'm not sure how to use Integration by Parts with the left side, since neither e^-x2 nor sin(2xt) gets simpler when differentiated.

 

[tiny-tim]

Differentiating the second one yields

e^-x2 * e^-x2 + ∫e^u2du * -2x²e^-x2 = 1 - 2xe^-x2∫e^u2du.

Yes (you got a minius wrong at the start, but you corrected it).

So if G(x) = e^-x2∫e^u2du, I have

G' = 1 - 2x * G
G' / G = 1 - 2x

No, G'/G is not 1 - 2x.

… Hmmmmm ... I'm not sure how to use Integration by Parts with the left side, since neither e^-x2 nor sin(2xt) gets simpler when differentiated.

You need to differentiate the first equation first … then use integration by parts.

 

[Jamin2112]

Yes (you got a minius wrong at the start, but you corrected it). No, G'/G is not 1 - 2x. You need to differentiate the first equation first … then use integration by parts.

Let me try again. (Wow! I feel like a moron for doing that algebra wrong.)

G' = 1 - 2x * G

... That's not separable. It's been a while since I took Intro To Differential Equations. How do I do this?

As for the left side, I have 2x∫e^-t2cos(2xt)dt. Unfortunately, this doesn't correspond to anything on the Laplace Transform table or Integral Table. You mentioned using Euler's thingy?

 

[Jamin2112]

Hang on. Let me do this again.

F'(x) = ∫-te^-t2sin(xt)dt.

Let u = sin(xt), dv = -te^-t2. Then we have

lim_t-->∞[ (-1/2)sin(xt)e^-t2 ] - x∫-(-1/2)e^-t2cos(xt)dr = (-1/2)xF(x).

To solve the resulting differential equation,
F' / F = -x/2
d/dx [ln(F)] = -x/2
ln(F) = -x²/4 + C
F = Ke^-x2/4


http://www.layoutcodez.net/personalized/google/success_baby70989908.jpg

 

[tiny-tim]

Hi Jamin2112!

(is that you on the beach, last summer? )

G' = 1 - 2x * G

... That's not separable. It's been a while since I took Intro To Differential Equations. How do I do this?

I don't think it can be done, except by using that theorem.

You mentioned using Euler's thingy?

hmm … I've changed my mind about that … I thought that replacing sin(2xt) by (e^2ixt - e^-2ixt)/2 would help, but it doesn't seem to.

Hang on. Let me do this again.

F'(x) = ∫-te^-t2sin(xt)dt.

No, that shoulld be a cos (and what happened to the 2xt ?)

Try again! ​

To solve the resulting differential equation,
F' / F = -x/2
d/dx [ln(F)] = -x/2
ln(F) = -x²/4 + C
F = Ke^-x2/4

uhh? i think you've been eating too many rusks!

 

[uart]

Let me try again. (Wow! I feel like a moron for doing that algebra wrong.)

G' = 1 - 2x * G

... That's not separable. It's been a while since I took Intro To Differential Equations. How do I do this?

As for the left side, I have 2x∫e^-t2cos(2xt)dt. Unfortunately, this doesn't correspond to anything on the Laplace Transform table or Integral Table. You mentioned using Euler's thingy?

Hi Jamin2112.

Yes getting the DE, G' = 1 - 2x * G, for the RHS is a good start.

You can get exactly the same DE for the LHS by first differentiating wrt x and then using integration by parts to rearrange it. I found this the easiest way to proof the identity. BTW. It's an interesting identity too :)

 

[uart]

Oh perhaps an even a better proof is to start out by generating the DE, G' = 1 - 2x * G for the LHS as outlined above, and then to solve it (using method of integrating factors) which leads explicitly to the RHS.