Show that # int_(-oo)^(oo) e^(-pix^2) \ dx = 1 #?
2 Answers
See below.
Explanation:
If
This integral is to be taken over the whole plane. Changing to polar coordinates we have
and then
We seek:
# I = int_(-oo)^(oo) e^(-pix^2) \ dx #
Which is a weighted variation of the Gaussian integral
# int_(-oo)^(oo) e^(-x^2) \ dx =sqrt(pi) #
This in itself is quite a remarkable result, and is fundamental in deriving probabilities for the Normal Distribution in Statistics.
We first note there is no elementary antiderivative for the function
Consider:
# I^2 = (int_(-oo)^(oo) e^(-pix^2) \ dx)^2 #
# \ \ \ \ = (int_(-oo)^(oo) e^(-pix^2) \ dx) \ (int_(-oo)^(oo) e^(-pix^2) \ dx) #
As we have definite integrals we can arbitrarily change the variable of integration, so let us do this for the latter integral:
# I^2 = (int_(-oo)^(oo) e^(-pix^2) \ dx) \ (int_(-oo)^(oo) e^(-piy^2) \ dy) #
Then using properties of double integrals, we can write this as:
# I^2 = int_(-oo)^(oo) \ int_(-oo)^(oo) \ e^(-pix^2) \ e^(-piy^2) \ dx \ dy #
# \ \ \ \ = int_(-oo)^(oo) \ int_(-oo)^(oo) \ e^(-pi(x^2+y^2)) \ dx \ dy #
If we look at the bound of integration in cartesian coordinates we are integrating over all of
#x# limits are# { (x = -oo), (x=oo) :} #
#y# limits are# { (y = oo), (y = -oo) :} #
So we can write integral as:
# I^2 = int int_(RR^2) \ e^(-pi(x^2+y^2)) \ dx \ dy #
And so if we switch to polar coordinates we can integrate over the same region by using the following bounds:
#r# limits are# { (r = 0), (r=oo) :} #
#theta# limits are# { (theta = 0), (theta = 2pi) :} #
Using :
# { (x=rcos theta), (y=rsin theta) :}=> x^2+y^2 = r^2 #
And, the Jacobian for polar coordinates is
So if we change to polar coordinates, we have:
# I^2 = int int_(RR^2) \ e^(-pi(x^2+y^2)) \ r \ dr \ d theta #
# \ \ \ \ = int_0^(2pi) int_o^(oo) \ e^(-pir^2) \ r \ dr \ d theta #
# \ \ \ \ = -1/(2pi) \ int_0^(2pi) int_o^(oo) \ (-2pi)e^(-pir^2) \ r \ dr \ d theta #
The inner integral is now trivial to integrate:
# I_("Inner") = int_o^(oo) \ (-2pi)e^(-pir^2) \ r \ dr#
# \ \ \ \ \ \ \ \ = [e^(-pir^2)]_0^(oo) #
Or strictly speaking , as this is an improper integral:
# I_("Inner") = lim_(n rarr oo) [e^(-pir^2)]_0^n #
# \ \ \ \ \ \ \ \ = lim_(n rarr oo) (e^(-pin^2) - e^0) #
# \ \ \ \ \ \ \ \ = 0 - 1 #
# \ \ \ \ \ \ \ \ = - 1 #
So then returning to our double integral we haver:
# I^2 = -1/(2pi) \ int_0^(2pi) \ I_("Inner") \ d theta #
# \ \ \ \ = -1/(2pi) \ int_0^(2pi) \ (-1) \ d theta #
# \ \ \ \ = -1/(2pi) [ - theta ]_0^(2pi) #
# \ \ \ \ = 1/(2pi) [ theta ]_0^(2pi) #
# \ \ \ \ = 1/(2pi) (2pi - 0) #
# \ \ \ \ = 1 #
And we require:
# I = sqrt(I^2) = 1 # QED
Where we take the positive value as