# Show that if equivalent to

Jamin2112
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

Homework Helper
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?

Homework Helper
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 + ∫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.

Homework Helper
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). So if G(x) = e-x2∫eu2du, 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

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

Homework Helper
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 (e2ixt - 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! 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 :)