Integration of Sin(x^2)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/ |
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