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/ |
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