Show that # int_(-oo)^(oo) e^(-pix^2) \ dx = 1 #?

2 Answers
Oct 3, 2017

See below.

Explanation:

If #I = int_(-oo)^oo e^(-pi x^2)dx# then

#I^2= (int_(-oo)^oo e^(-pi x^2)dx) (int_(-oo)^oo e^(-pi y^2)dy) =#

#int_(-oo)^oo int_(-oo)^oo e^(-pi(x^2+y^2))dxdy#

This integral is to be taken over the whole plane. Changing to polar coordinates we have

#x^2+y^2 = r^2# and

#dx dy = r dr d theta# and then

#I^2=int_(-oo)^oo int_(-oo)^oo e^(-pi(x^2+y^2))dxdy equiv int_0^(2pi)int_0^oo e^(-pi r^2) r dr d theta = 1#

and then

#I = int_(-oo)^oo e^(-pi x^2)dx=1#

Oct 3, 2017

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 #e^(-x^2)#, but by utilising some results from multivariable calculus along with a change of coordinate system we can readily evaluate the integral.

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 #RR# in the #x#-direction and similarly all of #RR# in the #y#-direction. In other words we integrating over the entire #RR^2# plane.

#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 #r \ dr \ d theta #

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 #e^x gt AA x in RR#, making the definite over #RR# positive.