Show that if equivalent to

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  • #1
Jamin2112
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Show that .... if equivalent to ....

Homework Statement



Show that

∫e-t2sin(2xt)dt (bounds: 0,∞)

= e-x2∫eu2du (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) = ∫eu2 du (bounds: 0, x). Then, using the fundamental theorem of calculus,

(e-x2G(x))' = e-x2 * ex2 + eu2 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.
 

Answers and Replies

  • #2
tiny-tim
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Hi Jamin2112! :smile:
∫e-t2sin(2xt)dt (bounds: 0,∞)

Wouldn't it be easier to use Euler's formula (cos + isin), and then complete the square ? :wink:
 
  • #3
Jamin2112
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Hi Jamin2112! :smile:


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

First of all, should I be differentiating both sides of the original equation?
 
  • #4
tiny-tim
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Hi Jamin2112! :smile:
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! :wink:), 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? :smile:
 
  • #5
Jamin2112
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Hi Jamin2112! :smile:


Yes.

Differentiate the second one first (so that you know what you're aiming at! :wink:), 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? :smile:



Differentiating the second one yields

e-x2 * e-x2 + ∫eu2du * -2x2e-x2 = 1 - 2xe-x2∫eu2du.

So if G(x) = e-x2∫eu2du, I have

G' = 1 - 2x * G
G' / G = 1 - 2x
d/dx[ln(G)] = 1 - 2x
ln(G) = x - x2 + C
G = ex - 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.
 
  • #6
tiny-tim
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Differentiating the second one yields

e-x2 * e-x2 + ∫eu2du * -2x2e-x2 = 1 - 2xe-x2∫eu2du.

Yes (you got a minius wrong at the start, but you corrected it). :smile:
So if G(x) = e-x2∫eu2du, I have

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

No, G'/G is not 1 - 2x. :redface:
… 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.
 
  • #7
Jamin2112
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Yes (you got a minius wrong at the start, but you corrected it). :smile:


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


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?
 
  • #8
Jamin2112
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Hang on. Let me do this again.

F'(x) = ∫-te-t2sin(xt)dt.

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

limt-->∞[ (-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
 
  • #9
tiny-tim
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Hi Jamin2112! :smile:

(is that you on the beach, last summer? :biggrin:)
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 (e2ixt - e-2ixt)/2 would help, but it doesn't seem to. :redface:
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 ?) :wink:

Try again! :smile:
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? :confused: i think you've been eating too many rusks! :biggrin:
 
  • #10
uart
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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 :)
 
  • #11
uart
Science Advisor
2,797
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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.
 

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