Ill you help me?"Unraveling the Mystery of 'e': Solving an Integral

[opeth_35]

actually, there ıs a something I don't understand about the e value

I could not figure it out this integral while solving for quantum physics, you will see the equation in figure. The problem is how We can make a solution about the e number while solving in integral.. Such as e power any number.. you wıll see in figure as well. I just wonder only how to solve problem which contain e number. I knew before about that, W

 

[gomunkul51]

Excuse me, but I don't understand your question.
Please rephrase it, or write it mathematically and mark what you need to solve.

good luck.

 

[tiny-tim]

hi opeth_35!

i too don't understand what you're asking

if you're asking how to solve ∫_-∞^∞ ue^-λu2 du and ∫_-∞^∞ e^-λu2 du, the first one is by substitution and the second one is a well-known definite integral which you have to learn

 

[opeth_35]

I could not complete my question sorry about that I think there is a problem about web. anyway.

If I tell my questıon to you again.. I don't understand how to calculate contain e number ( 2.7 )

this kind of integral questions. You wıll see in mpeg the sample of any question relevant that.

Is there anyone know about how i m going to solve this ?

 

[tiny-tim]

I could not complete my question sorry about that I think there is a problem about web. anyway.

If I tell my questıon to you again.. I don't understand how to calculate contain e number ( 2.7 )

this kind of integral questions. You wıll see in mpeg the sample of any question relevant that.

Is there anyone know about how i m going to solve this ?

hi opeth_35!

i'm sorry, i still don't understand what you're asking

e is just like any other constant, the only difference is that d/dx (e^f(x)) = f'(x)e^f(x), while if we use a different number, A, instead of e, then d/dx (A^f(x)) = (ln(A))f'(x)A^f(x), which is more messy


( and ∫ f'(x)e^f(x) dx = e^f(x), while ∫ f'(x)A^f(x) dx = A^f(x)/ln(A) )

 

[gomunkul51]

@opeth_35 I advise you to write your question in the language of mathematics.

 

[opeth_35]

:) :) I thınk I could not explain problem clearly.. I understand that added solve of questıon in mpeg but

when we take integral, then We need to put boundaries from minus infinite to plus infinite. Am I right? Actually, This point confused me.. I am putting minus infinite and plus infinite to solve but The answer is not like I found. I think I forgot something relevant e number how to solve this kind of boundries. Am I clear? :)

 

[tiny-tim]

hi opeth_35!

the important values are e⁰ = 1, and e^-∞ = 0

in this case you have both ∞ and -∞, but it doesn't matter because they're both squared, so at both boundaries it's e^-(∞)2, or e^-(-∞)2, which are both e^-∞ = 0

 

[gomunkul51]

like tiny-tim said :

e^-(-∞)2 = e^-(∞)2= e^-∞ = 1/e^∞ = 1/∞ = 0

:)

 

[opeth_35]

yes all of you are right that you showed me, the best way, I think It is going to be show you my solution..:)

Could you check a photo I added here?

I hope This will be clear. Please help

 

[tiny-tim]

i'm sorry, opeth_35, this is completely wrong …

∫ e^-2ax2 dx is not e^-2ax2/(-4ax) …

(differentiate the RHS, and you'll se you can't possibly get the LHS)

there is no elementary integral of e^-2ax2 (though of course there is the "non-elementary" error function, erf(x), see http://en.wikipedia.org/wiki/Error_function" )

to integrate ∫₀^∞ e^-2ax2 dx, you need to learn a standard trick:

multiply it by ∫₀^∞ e^-2ay2 dy to get an ∫∫ dxdy, then convert to polar coordinates …

what do you get? ​

 

[opeth_35]

thank you for your propose :) I did not know before this kind of questions that was solved like that..

I will try to solve like you said :)

 

[opeth_35]

I multiplied dy function as you said but I can not move in progress. I see again similar wrong in process.. I am confused:(

 

[tiny-tim]

show us your full calculations, and then we'll see what went wrong!

(btw, I'm afraid I'm going out in a few minutes, and i'll be out for the rest of the evening)

 

[SammyS]

Your 1st image shows:

$$\left\langle x\right\rangle=A\int_{-\infty}^\infty xe^{-\lambda(x-a)^2}\,dx=\dots=A\left(0+a\sqrt{\frac{\pi}{\lambda}}\right)=a$$

It's hard to figure what your question is even after reading all the posts in this thread.

I take it that A is a normalization constant.

Indeed, $$A\int_{-\infty}^\infty xe^{-\lambda(x-a)^2}\,dx=A\,\sqrt{\frac{\pi}{\lambda}}\quad\to\quad A=\sqrt{\frac{\lambda}{\pi}}\,,$$ which explains: $$A\left(0+a\sqrt{\frac{\pi}{\lambda}}\right)=a\ .$$

If on the other hand you wonder how $$\int_{-\infty}^\infty e^{-x^2}\, dx=\sqrt{\pi}\,,$$ that's more involved.

 

[tiny-tim]

SammyS, whatever are you doing?

You've just given him the complete answer.

 

[SammyS]

SammyS, whatever are you doing?

You've just given him the complete answer.

Sorry Tim.

I was just trying to figure out the question. Let's see if it's not too late to edit.

FIXED it !

Tim, Thanks!

 

[tiny-tim]

No harm done!

 

[opeth_35]

okey, I understand to solve but I have forgotten something in math.. I need to make a reputation about coordinate system..

thank you all:)

 

[tiny-tim]

hey

I think My problem is about coordinate systems, I still don't understand to turn -infinite to the number of pi..

Do u have any suggestıon about that:(

once again I'm not understanding you

show us the particular equation you're having difficulty with

 

[opeth_35]

You ll see in mpeg..

 

[tiny-tim]

(jpeg, actually! )

ok, now put e^-2ax2e^-2ay2

= e^-2a(x2+y2)

= e^-2ar2 …

finally convert "dxdy" to polar coordinates ​

 

[opeth_35]

here we go, My problem is this.. polar koordinates.. I can not turn to each other. I have looked to book but, It ıs not clear.. It explains roughly..:(

 

[tiny-tim]

you have to replace dxdy by rdrdθ …

see for example http://www.ltcconline.net/greenl/courses/202/multipleIntegration/doublePolarIntegration.htm"

 

[SammyS]

you have to replace dxdy by rdrdθ …

see for example http://www.ltcconline.net/greenl/courses/202/multipleIntegration/doublePolarIntegration.htm"

... and then figure out the limits of integration in polar coordinates --- so that you integrate over the whole coordinate plane.

 

[opeth_35]

hey.. I stucked again in polar coordinates, any help?

 

[tiny-tim]

hi opeth_35!

∫_-∞^∞∫_-∞^∞ dxdy

is over all of x and y,

so in polar coordinates, that's all of r and θ, ie the limits will be 0≤r≤∞, 0≤θ≤2π

 

[opeth_35]

how do u know, the boundries for teta between 0 and 2pi , haven't figure it out?

and why is going from 0 to -ınfınıte for r, It could be from -infinite to +infinite ?

 

[tiny-tim]

no, in polar coordinates, r cannot be negative, it is defined as being 0≤r<∞ …

likewise, θ is defined as being 0≤θ<2π

 

[opeth_35]

:) finally I found one value for solve , but the book shows different value, Could u check my calculate?

 

[tiny-tim]

ah, you've calculated 4 times ∫₀^∞∫₀^∞ dxdy …

that's 4 times ∫₀^∞∫₀^π/2 drdθ, isn't it?

 

[opeth_35]

I think, I will not change the function when I calculate..

 

[tiny-tim]

sorry, you're misunderstanding …

the "4" is ok

it's the "2π" in your limit which was wrong, and needed to be changed to "π/2" (in red in my last post)

 

[opeth_35]

so, why did the function go from 0 to pi/2 ? :(

I don't understand that, on the other hand, the function could go from 0 to 2pi...

when can we see this kind of problems because I haven't figured out :(

 

[tiny-tim]

because your ∫∫ from 0 to ∞ was only over the first quadrant (x≥0, y≥0) …

that's 0 ≤ θ < π/2 ​

(if you'd kept to ∫∫ from -∞ to ∞, ie over all four quadrants, that would be 0 ≤ θ < 2π)