#1
Ok guys I'm doing my take-home Advanced Quantum Mechanics test and I need some help on an integral....I'm gonna use the "{" bracket for integrals

Ok, so I have {[x^6][e^(-x^2)]dx

EDIT: this is an indefinite integral, i.e. from negative infinity to positive infinity.

I know that this has to reduce to 15[sqrt(pi)]/8 because I have the answer in my notes lol, but when we went over this in class we skipped the integration to save time; this is also in the book so I know it's correct (the book also skips the integration...)

The only hint I have is that {[e^(-x^2)]dx = sqrt(pi). I've tried and tried but just can't get this to work out. Any ideas? Thanks a bunch
Last edited by rcw110131 at Apr 29, 2008,
#3
Integration by parts dude. Or table method. When all else fails, use Maple 10 (computer calculator that can do anything you can dream of.

EDIT : do you have integration limits?
Last edited by Trisonic77 at Apr 29, 2008,
#4
MAple tis' quite amazing.
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#5
Use integration by parts. I'll outline it for you.

u = x^6
du = 6x^5 dx

v = e^(-x/2)
dv = -2e^(-x/2)

Integration by parts works by the following formula:

integral(f(x)) (where u and v are your separate functions) = uv - integral(du*v)

So by that formula, you have

integral([x^6][e^(-x^2)]dx) = (x^6)(e^(-x/2)) - integral( (6x^5)*(-2e^(-x/2)) dx)

Now repeat the integration by parts five more times until the "u," namely the x^n term reduces to a constant.
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#6
I have no idea how to integrate the e^(-X^2), but if you know what the integral of it is, you can just use integration by parts.

Take x^6 as u, take e^(-X^2) as dv/dx. word out v and du/dx.

Then put them into the formula uv-{(v)(du/dx) dx. It looks like you may have to go through that a huge number of times though.
#7
hah..I have a test on this on thursday...damn. totally random that its up here cause my class tonight was doin this..wEirD!!
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#8
Quote by SmarterChild
It looks like you may have to go through that a huge number of times though.


Six, to be exact. It works out nicely (and much faster) to use an IBP table though...
Quote by denizenz
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#9
i think you can use a reduction formula or split it into two
X^5 * xe^(x^2) so the exponential part you can integrate by recognition and thus do it by parts. The answer comes to root pi something or other if your limits are from infinity to minus infinity
#10
Quote by darkstar2466
Use integration by parts. I'll outline it for you.

u = x^6
du = 6x^5 dx

v = e^(-x/2)
dv = -2e^(-x/2)

Integration by parts works by the following formula:

integral(f(x)) (where u and v are your separate functions) = uv - integral(du*v)

So by that formula, you have

integral([x^6][e^(-x^2)]dx) = (x^6)(e^(-x/2)) - integral( (6x^5)*(-2e^(-x/2)) dx)

Now repeat the integration by parts five more times until the "u," namely the x^n term reduces to a constant.


Yeah I tried integration by parts already but the "v" is e^(-x^2), not e^(-x/2). And by the way IBP is int[udv] = uv - int[vdu], but you prolly just made a typo lol.

And as I said before, integral[e^(-x^2)] reduces down to just the square root of pi (the actual hint integral given is integral[e^(-cx^2)] = sqrt(pi/c), but c=1 in this case).
#12
ya if you do parts you get

u=x^6 dv=e^(-x^2)dx
du=6x^5dx v=sqrt(pi) <--from the hint you got

which by the formula comes out

x^6*sqrt(pi) - integral[6x^5*sqrt(pi)dx]

and since sqrt(pi) is just a number it can come outside the integral and be factored out but i'm not sure how they get the final answer you have in your notes because as best as i can see unless there is something stupid i am doing i get

sqrt(pi)[x^6 - x^6].... after the integration of 6x^5 and factoring out sqrt(pi)
and that gives you a grand total of 0... so i have no idea how you are supposed to get that answer because i think i did this right
Last edited by stubs_88 at Apr 29, 2008,
#13
Quote by rcw110131
Yeah I tried integration by parts already but the "v" is e^(-x^2), not e^(-x/2). And by the way IBP is int[udv] = uv - int[vdu], but you prolly just made a typo lol.

And as I said before, integral[e^(-x^2)] reduces down to just the square root of pi (the actual hint integral given is integral[e^(-cx^2)] = sqrt(pi/c), but c=1 in this case).


Oh boy. Not only did I misread the question, but I also made a typo. Long night...
Quote by denizenz
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#14
Quote by stubs_88
ya if you do parts you get

u=x^6 v=e^(-x^2)dx
du=6x^5dx dv=sqrt(pi) <--from the hint you got

which by the formula comes out

x^6*sqrt(pi) - integral[6x^5*sqrt(pi)dx]

and since sqrt(pi) is just a number it can come outside the integral and be factored out but i'm not sure how they get the final answer you have in your notes because as best as i can see unless there is something stupid i am doing i get

sqrt(pi)[x^6 - x^6].... after the integration of 6x^5 and factoring out sqrt(pi)
and that gives you a grand total of 0... so i have no idea how you are supposed to get that answer because i think i did this right


Yep I just did the same thing and got that too....

I have in my notes that integral[e^(-cx^2)dx] = sqrt[pi/c] and that we're differentiating with respect to c and treating x as constant (in my problem c is just 1). Then I jotted down "repeat 3 times", but unfortunately I'm pretty sure I didn't understand what he meant when I wrote this down, nor now lol.
#15
What in the hell is going on here? What kind of math is that?
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#16
Quote by dannyniceboy
What in the hell is going on here? What kind of math is that?


Integral Calculus.
Quote by denizenz
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#17
**** I probably should've mentioned that my limits are from negative infinity to infinity....
#19
i'm not going to bother deriving the reduction formula but here it is (n is the power that x is raised to)




I think you made several mistakes in copying because the answer appears to be (15sqrt[pi])/16 not 15[sqrt(pi)]/8 and you copied the hint wrong as well.... (what the ****) the integral of e^(-x^2) is sqrt(pi)/2,

Quote by seljer
be lazy and throw it into mathematica?

where the Erf is http://en.wikipedia.org/wiki/Error_function

if you read his post, he needs help with solving the problem... spitting out an answer using CAS like Mathematica is not going to help come test time
Last edited by Negative Burn at Apr 29, 2008,
#20
^According to Mathematica and my TI-84, the integral of e^(-x^2) is indeed just the square root of pi, not the square root of pi over 2.

Ugh I'm so frustrated! I really appreciate all the help though from everyone, thanks a bunch. Hopefully we'll figure this out before I have to leave for class at 10am! Lol.
#21
Quote by Negative Burn
if you read his post, he needs help with solving the problem... spitting out an answer using CAS like Mathematica is not going to help come test time


yeah, i know, i just love it because I've gotten a ton of beer using it to solve other freshman's homework
#22
Quote by seljer
yeah, i know, i just love it because I've gotten a ton of beer using it to solve other freshman's homework

rofl

Quote by rcw110131
^According to Mathematica and my TI-84, the integral of e^(-x^2) is indeed just the square root of pi, not the square root of pi over 2.

According to Mathematica's Integrator: http://integrals.wolfram.com the integral of e^(-x^2) is indeed sqrt(pi)/2. Try it yourself. Also, Maple says the same thing. And another source says the same thing as well:



Trust me yet?

Just memorize the reduction formula I gave you and remember the integral of xe^(-x^2) is 1/2, while the integral of e^(-x^2) is sqrt(pi)/2. With that information and the reduction formula you can solve any integral in the form of x^(n) e^(-x^2) dx where n => 2.

I'm going to bed (it's 5AM here). Good luck on the exam dude
Last edited by Negative Burn at Apr 29, 2008,
#23
I'm sure i remember having this problem thrown at me once, and i'm pretty sure there's a substitution or a trick that will solve it much more easily than by parts or reduction...only problem is, i can't remember the damn thing... it definately involved the sqrt(pi), but beyond that, i can't remember.
#24
^ According to the Integrator it's sqrt(pi)/2 * erf(x), and evaluated from negative infinity to positive infinity, I'm pretty erf(x) --> 2, but I don't know for sure.

Also this, is in my book, I know that 15sqrt(pi)/8 is the correct answer. Basically the problem here is the first order perturbation energy of an anharmonic oscillator, and it goes from
2/sqrt(pi) * [bunch of constants] * int[(x^6)(e^(-x^2))dx]

to

15/4 * [same bunch of constants].

By this you can figure out that the integral HAS TO reduce to 15sqrt(pi)/8.
#25
EDIT: Okay. Regardless of the value it works out using that reduction formula. I'm sure there is an easier/faster way, but so far no has come up with better--SO JUST USE IT lol



Simple, ain't it?
Last edited by Negative Burn at Apr 29, 2008,
#26
There are simple ways of doing by integrating over the complex plane. Also once you get the value of the e^(-x^2) integral, you can multiply the argument by an arbitrary constant a, so that you have e^(-ax^2) and then differetiate under the integral sign until you get an x^6 multiplying term.

The brunt force method of doing this integral is to keep on doing integration by parts until you get the integral of e^(-x^2), which is a Gaussian from -infinity to + infinity. It has a pretty well known value of sqrt(pi), but to do the integral you can consider taking the integral over the entire xy plane, and then converting to polar coordinates. The integral will be the square root of that answer. It's outlined here:

http://mathworld.wolfram.com/GaussianIntegral.html

If it's a physics test though, I doubt you need to go through all that, just quote the result of the Gaussian at the end of integration by parts.