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Convergence of Cos(x^2) or Sin(x^2)


Date: 11/15/97 at 09:14:52
From: Richard Underwood
Subject: Convergence of cos(x^2) or sin(x^2)

How would you prove that the integral from n=0 to infinite of 
sin(x^2)dx or cos(x^2)dx converges?  

Thank you very much; this will be a great help.

Sincerely,
Richard Underwood


Date: 11/15/97 at 11:10:11
From: Doctor Anthony
Subject: Re: Convergence of cos(x^2) or sin(x^2)

The Fresnel integrals are:

INT(0 to infinity)[cos(x^2).dx] = INT(0 to infinity[sin(x^2).dx]

                                = (pi/8)^(1/2)

This result can be obtained directly from the integral of e^(-x^2) 
from 0 to infinity, and using cos(x^2) = sin(pi/2-x^2)

                            = Imaginary part of  e^(i.(pi/2 - x^2))
                            
                            = Im. e^(i.pi/2) e^(-i.x^2)

The following shows how we integrate e^(-x^2/2), the normal 
distribution curve.

INTEGRATING THE NORMAL CURVE

I will carry out the integral from 0 to infinity, and doubling the 
result (since the graph is symmetrical about the y axis) will give the 
total area from -infinity to +infinity.

 Let  I = INT(0 to infinity)[e^(-x^2/2).dx]

This cannot be evaluated using elementary methods, so we proceed as 
follows:

  I^2 = [INT(0 to infinity)(e^(-x^2/2).dx]^2

  = INT(0 to infinity){e^(-x^2/2)dx) INT(0 to infinity)(e^(-y^2/2).dy)

  = INT(dx INT[e^(-(x^2+y^2)/2).dy]   all integrals are from 0 to 
                                      infinity.

  = INT.INT[e^(-(x^2+y^2)/2).dx.dy]

where the region of integration is the whole of the positive quadrant 
of the xy plane.

If we transform to polar coordinates, x^2+y^2 = r^2, also the element 
of area dx.dy is now given in polar coordinates by element of area 
r.d(theta).dr  The limits of integration will be 0 to infinity for r, 
and 0 to pi/2 for theta.

 So our integral now becomes

  I^2 = INT.INT[e^(-r^2/2)r.dr.d(theta)]

      = INT[d(theta)INT[e^(-r^2/2)r.dr]

make the substitution  r^2/2 = u    then r.dr = du and the inner 
integral becomes

           INT[e^(-u).du]   for u from 0 to infinity.

           = - e^(-u)  = -(0 - 1) = 1

So now we have  

   I^2 = INT[d(theta)] from 0 to pi/2

       =  [theta] from 0 to pi/2

       =  pi/2

and so I = sqrt(pi/2)

and 2I = sqrt(2pi)  giving area from -infinity to +infinity.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 11/15/97 at 13:28:42
From: Doctor Jerry
Subject: Re: Convergence of cos(x^2) or sin(x^2)

A more primitive approach to this problem is to first consider 
the integral int(a,b,sin(x^2)dx). Make the substitution w=x^2. 
Integrate the resulting integral by parts, letting u = w^(-1/2) 
and dv = sin(w)dw.  This will give

int(a,b,sin(x^2)dx)=
   (1/2)[-cos(b^2)/b+cos(a^2)/a-(1/2)int(a^2,b^2,cos(w)*w^(-3/2)*dw)]

Using easy estimates (including replacing cos(w) by -1), this gives

|int(a,b,sin(x^2)dx)|<=(1/2)(1/a+1/b+(1/2)(2/a-2/b))=1/a.

Then, going back to the original integral,

 int(0,b,*)=int(0,sqrt(pi),*)+int(sqrt(pi),sqrt(2pi),*)+...
     +int(sqrt((n-1)pi,sqrt(n*pi),*)+int(sqrt(n*pi,b,*).

So, the behavior of the original integral is the behavior of a 
convergent alternating series and a term that (from the above 
estimates) goes to zero.

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus

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