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### Integration of Sin(x^2)

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Date: 11/10/97 at 00:14:46
From: Joshua
Subject: Integration of Sin(x^2)

Hello, Doctor! For the last couple of months I have been occasionally
working on integrating Sin of x^2. Being that I have no formal
calculus training past basic integration, I have found it quite
difficult (most of my subsequent calculus knowledge has been
intuitive; that's the fun part :) . I have been told that it has no
finite solution, which I concluded myself, I mean, look at the
equation, right? But my question remains! I have been given the
solution in the form of Frensel's Sin, which is all fine, but it
explains nothing about how it was integrated. I am not looking for an
equation, I am looking for a reason! If you could guide me at all in
this quest, or point me to a book or html, I would greatly appreciate
it. Thanks.

Josh, the patient
```

```
Date: 11/10/97 at 12:58:17
From: Doctor Anthony
Subject: Re: Integration of Sin(x^2)

You can express sin(x) as an infinite series

sin(x) = x - x^3/3! + x^5/5! - x^7/7! + .....  and so

sin(x^2) = x^2 - x^6/3! + x^10/5! - x^14/7! + ......

Clearly you can take as many terms as you like to achieve the required
accuracy. You then integrate term by term.

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, the element of
area dx.dy is also 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/
```
Associated Topics:
College Calculus
College Trigonometry
High School Calculus
High School Trigonometry

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