Question

I have a problem understanding converting the integral of x^2 exp(ax^2) into polar form.

Answer 1

Could anyone explain it?

Answer 2

As much as I understand:\[I=\int\limits_{-\infty}^{\infty}x^2 e^{ax^2}dx\]
\[I^2=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}x^2 e^{a(x^2+y^2)}dxdy\]
Fro, here I fail to understand.

Answer 3

*m

Answer 4

Or even where the first x^2 is just x^1, for simplicity.

Answer 5

that integral u wrote up there diverges ...

Answer 6

Or from 0 to infinity, my problem here isn't the exact integral, but how you convert it into polar, and why.

Answer 7

the only integral in that form i know is\[I=\int\limits_{-\infty}^{\infty} e^{-ax^2}dx\]with \(a>0\)

Answer 8

Fine, then- just how do you convert it into polar (I can't remember the exact problem, sorry)?

Answer 9

np henpen\[I^2=\int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} e^{-a(x^2+y^2)} \text{d}x \ \text{d}y\]so i think no problem till here

Answer 10

Yes. I know that \[y^2+x^2=r^2=q\], (but whether r or q we will use here I do not know). Do we just substitute?

Answer 11

yeah integrating \(f(x,y)\) in cartesian with dxdy equals to integrating \(f(r,\theta)\) in polar with rdrdtheta and of course we must change integrals bounds

Answer 12

So it's not a case of robotically substituting and inserting the new values for dx, dy, y and x?

Answer 13

this is like a doing sub but with 2 variable

Answer 14

I get the intuition, but I still feel shaky. What is the proof of this (or, what f(x,y) is equal to an g(r, theta))?

Answer 15

I mean, with substituting, r=x^2+y^2, dx=dr/2x, dy=dr/2y. Where does theta fit in to this?

Answer 16

my friend herp-derp will clarify us

Answer 17

Very kind of you, herp-derp.

Answer 18

\[\text{If}~~I=\int_0^\infty x^2 e^{-ax^2}\text{d}x\]\[\text{then}~~I^2=\left(\int_0^\infty x^2 e^{-ax^2}\text{d}x\right)\left(\int_0^\infty y^2 e^{-ay^2}\text{d}y\right)=\int_0^\infty\int_0^\infty x^2y^2e^{-a(x^2+y^2)}\,\text{d}x\]You wrote it as \(\int_0^\infty\int_0^\infty x^2 e^{-a(x^2+y^2)}\text{d}x\,\text{d}y\) . Just wanted to clarify!

Answer 19

Sorry, not really answering your question...

Answer 20

Is that a dy missing from the end of your long equation?

Answer 21

Yes.

Answer 22

I doubt it's a straight substitution, but what's the general rule for where you can/can't polarly substitute (and how)?

Answer 23

It's here http://en.wikipedia.org/wiki/Polar_coordinate_system#Generalization , but I don't understand all of it.

Answer 24

http://tutorial.math.lamar.edu/Classes/CalcIII/DIPolarCoords.aspx

depends on our integral...sometimes changing to polar will make the integral very easy to solve ...sometimes no... there is no general rule

Answer 25

There must be a rule for conversion. I'll read the link.

Answer 26

for the particular case of\[I^2=\int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} e^{-a(x^2+y^2)} \text{d}x \ \text{d}y\]thee integral becomes to\[I^2=\int\limits_{0}^{2\pi} \int\limits_{0}^{\infty} e^{-ar^2} r\ \text{d}r \ \text{d}\theta\]which is very easy to solve because of the 'r' that changing to polar provides

Answer 27

People - get awareness finally that the x^2 factor at the start of the integrand equals\[(r* Cos \theta)^2\]

Answer 28

Soo overall u obtain
\[\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}r^2 Cos^2 \theta \exp^{-\alpha r^2} r dr d \theta\]

Answer 29

However Mr. H.D. here gave a smart and unobvious path to the solution itself

Answer 30

However his path MUST cite atheorem allowing the double integral to equal a product of usual integrals

Answer 31

henpen, for your case it is best to use a technique called "differentiation through the integral", because a change to polar coordinates will just make this integral messy.

So notice that \(\displaystyle \int_0^\infty x^2e^{-ax^2}\,\text{d}x\) is equivalent to \(\displaystyle\int_0^\infty -\frac{\partial}{\partial a}\left[e^{-ax^2}\right]\,\text{d}x\). Using our knowledge of differentiation through the integral, we know that the last expression is equal to \(\displaystyle -\frac{\partial}{\partial a}\int_0^\infty e^{-ax^2}\,\text{d}x\). So what we need is that integral.

Well, I will not explain it, but using a change to polar coordinates we can show that \(\displaystyle\int_0^\infty e^{-ax^2}\,\text{d}x=\sqrt{\frac{\pi}{4a}}\). Thus, \(\displaystyle -\frac{\partial}{\partial a}\int_0^\infty e^{-ax^2}\,\text{d}x=\frac{1}{4}\sqrt{\pi}a^{-3/2}\), which is your answer.

Answer 32

Hey You knew ! You knew! I am betting you read this BRILLIANT move somwh. !

Answer 33

Oh, I made a mistake. The derivative operator on the outside of the integral is not a partial derivative. It should read \(\displaystyle \frac{\text{d}}{\text{d}a}\int_0^\infty e^{-ax^2}\,\text{d}x\). Not that it matters.

Answer 34

@Mikael Tis true, I did not come up with this method by myself. \(:(\)

Answer 35

Genius method!

Answer 36

@henpen Do you understand differentiation through the integral?

Answer 37

First - give him another medal deserves so
Second Her Derp - u must know som probability distrib or physics

Answer 38

Yes, it's really elegant!

Answer 39

Integrals of the form:\[\int_a^b \frac{\partial^{\,n}}{\partial^{\,n} x}\Big[f(x,\xi)\Big]\,\text{d}\xi=\frac{\text{d}^n}{\text{d}^nx}\int_a^b f(x,\xi)\,\text{d}\xi\]

Answer 40

Basically, my original question is answered in that dA=rdrdtheta in general. Thanks for the link, mukshala!

Answer 41

Why does it switch from partial to 'normal'?

Answer 42

anytime henpen

Answer 43

And if it's from a to b (i.e. definite), surely the right hand side is =0

Answer 44

\(\displaystyle\int_a^b f(x,\xi)\,\text{d}\xi\) is \(\underline{\mathbf{not}}\) a function of \(\xi\). However, it \(\underline{\mathbf{is}}\) a function of \(x\).

Answer 45

And sorry, mukshala, I think you had actually explained this to me before, but I forgot...

Answer 46

Or- can the bounds be written as\[\xi(a)and \xi(b)\]? I think I get it.

Answer 47

Imagine a 3-dimensional surface defined by \(z=f(x,y)\). Imagine a strip on the x-y plane from y=a to y=b. Now imagine a cross-section of this surface parallel to the x-z plane. This cross-section forms a curve. Finally, imagine finding the area under that curve. As you move the cross-section, the curve changes, and so does its area. In such a way, the area of the curve depends on the x-coordinate of the cross section. It doesn't depend on any y or z-coordinate.

Answer 48

I did an example, and now I undertand that it works. You are an artful explainer, thank you a lot.

Answer 49

I went to an art school for 3 years. (It was hell)