We all know that the integral of exp(-x^2) dx from -oo to oo is sqrt(Pi). The derivation of this integral is especially simple if we find the square of the integral then use a double integral with polar coordinates (in any calculus book). However, I am interested in deriving
integral exp(-x^2) dx from -oo to oo without using that method (i.e. double integral, polar coordinates, more than one variable, erf function). I was trying power series and I got something that looked indeterminate. Any ideas (the simpler the better)?
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I have discovered a way to do this, except it involves the identity Gamma(x)*Gamma(1-x) = PI/Sin(PI*x). Can anyone prove this without double integrals? I saw a proof with contour integration but that is a little difficult to understand. Also appreciated is if anyone could explain that proof to me. Thanks!
lj8866
>>integral exp(-x^2) dx from -oo to oo
-let x^2=t (changing variables)
then work on la place transform of exp(-t). and with changing the limits (because we changed the variable)you will be able to get the same answer (we used this method in solving electrical circuits)
-another methode is to get the Taylor or McLauren series for exp(-x^2), you will take as many elements in order to let your error to be smalland you integrate each element + the error you shoosed you will get the exact answer...
sorry for not writing the solution because it will be unreadable ( exp(x^2...), very looonnnnggg in a non math platform)
NOTE: using polar coordinates will be much easer
sorry for my bad english
trinety
>>integral exp(-x^2) dx from -oo to oo
-let x^2=t (changing variables)
then work on la place transform of exp(-t). and with changing the limits (because we changed the variable)you will be able to get the same answer (we used this method in solving electrical circuits)
Can you show me how you would solve that with that particular transform?
-another methode is to get the Taylor or McLauren series for exp(-x^2), you will take as many elements in order to let your error to be smalland you integrate each element + the error you shoosed you will get the exact answer...
if you do that, you get a predefined function called the error function (erf(x)) and that is defined to have the properties of that integral.
And what do you mean by polar coordinates?
sorry for being late to replay but a storm cut off the communication lines...
i hope that this website would help you ..
http://mathworld.wolfram.c
best regards
trinety
the ame of all methods is to try to express this function >>exp(-x^2)<< in a way that we can do its integrals.
and because of my bad English i don't know how to explain it to you..
all i can say is that Taylor or McLaurin series will do the job because they transform any function into a series of powers of X ..
and i can't explain more about those series, you will learn them in higher levels of education.
sorry for any incoviennient
trinety
NOTE : bcz you did not got the answer for your question...
>> DO NOT ACCEPT ANY OF COMMENTS AS AN ANSWER <<
thank you
to solve this problem in the easiest way .. you have to know that dxdy in chartizian is rdrd(teta) in polar, due to the Jacobian transform....
some notations: (to simplify in this none math platform)
# exp(...) is e(...)
# d(teta) is dt
# +infinity = ->
# -infinity = <-
# integral from 0 to -> = int(0,->)(...)
# sqaure root of(...)= sqt(....)
back to our problem :
f(x)=e(-x^2) is the same as F(x,y)=e(-(x^2+y^2))
because it a little change of variable... both -x^2 and -x^2-y^2 are negative ...
and insted of studying from -infinity(<-) to +infinety(->) we will study it from 0 to -> bcz F(x,y)=F(-x,-y) (symetry about y axes)
not that x=r cos(t) ...E1
y=r sin(t) ...E2
wich gives us x^2+y^2 = r^2
int(0,->)(int(0,->)(e(-(x^
int(0,pi/2)(int(0->)(r e(-r^2)dr)dt)
we read the following statement as follows :
intgral1 from 0 to -> of (integral2 from 0 to -> of ( e(-(x^2+y^2))dx)dy)= integral1 from 0 tp pi/2 of (integral2 from 0 to -> of(r e(-r^2)dr)dt)
we changed the limits due to the change of variable (E1 & E2)
we do the first integral inside wich is now very easy
int(0,->)(r e(-r^2)dr) by tabular or by parts or any method you want ...
you get 1/2 as answer
now you still have int(0,pi/2)(1/2 dt)
(integral from 0 to pi/2 of(1/2 dt))
you will get pi/4 as answer...
we have not yet reched the final answer...
let I = int(0,->)(e(-x^2)dx)=int(0
I= integral from 0 to -> of e(-x^2) = integral from 0 to -> of e(-y^2)
actualy they are the same or with x or with y just any two variables
calculating I^2= (int(0,->)( e(-x^2)dx)* (int(0,->)( e(-y^2)dy)=
int(0,->)(int(0,->)(e(-(x^
witch we calculated = pi/4 it nothing then I^2=pi/4 then
I=sqt(pi/4)=(sqt(pi))/2
(note that "I" can't be negative bcz it is an area that we are calculating, double integrals)
don't forget that we were calculating half of our integral from 0 to -> and we said that it is symetric then the integral from <- to -> is nothing then double the integral from 0 to->
then sqt(pi/4)*2 will give us sqt(pi)...
see what i ment before that this is very difficulte to write in a none math format ...
sorry for any disadvantages and beeing rood to you ...
any question we are here to help
trinety
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by: GwynforWebPosted on 2003-02-06 at 12:46:01ID: 7896673
lj8866,
There are a number of "different" ways of proving this. All the ones I have seen are in fact the same as the proof you described but using a different notation. For instance I was taught the integral using complex analysis but it is really the same proof.
If anyone comes up with one that is truly different I would like to see it my self.
Gwynforweb