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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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