My Math Forum Transforming Integral

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 April 20th, 2017, 12:19 PM #1 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Transforming Integral I understand we can transform the following integral, by substituting y=0.25x, i.e dx = 4dy, x^2 = 16y^2 $\displaystyle \int_{0}^{4} \sin (x^2) dx = 4\int_{0}^{1} \sin (16y^2) dy$ But I don't understand how the following integral can be transformed to an integral with limits 0 and 1 $\displaystyle \int_{0}^{\infty} \sin (x^2) dx$ The answer states it assumes the integral converges, and that you can transform it using the transform y = (1/x) + 1. What does it mean to say an integral converges, and if I didn't have that answer how could I calculate what the transform should be? Last edited by skipjack; April 20th, 2017 at 02:25 PM.
 April 20th, 2017, 02:39 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,951 Thanks: 1599 If you substitute y = 1/x, what are the values of y that correspond to the given limits for x?
 April 22nd, 2017, 02:58 AM #3 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Ok, if I use the transformation y = 1/(x+1) x = (1/y) - 1 x = infinity, y = 0 x = 0, y =1 $\displaystyle \frac{dy}{dx} = \frac{-1}{(x+1)^2}$ $\displaystyle -(x+1)^2 dy = dx$ $\displaystyle -(1/y)^2 dy = dx$ $\displaystyle \int_{0}^{\infty} \sin (x^2) dx = \int_{0}^{1} -(1/y)^2 \sin (((1/y)-1)^2) dy$ I tried to check whether integrals evaluated to the same on Online Integral Calculator • Shows All Steps!. $\displaystyle \int_{0}^{\infty} \sin (x^2) dx ~= 0.6266$ However, it couldn't find an approximate answer for $\displaystyle \int_{0}^{1} -(1/y)^2 \sin (((1/y)-1)^2) dy$ Does my calculation look right? Last edited by skipjack; April 23rd, 2017 at 12:16 AM.
 April 22nd, 2017, 08:15 AM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 1,930 Thanks: 998 This is a Fresnel integral How to solve
 April 22nd, 2017, 08:52 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,805 Thanks: 1045 Math Focus: Elementary mathematics and beyond Hi calypso. Your work is good except for a sign error. The limits in terms of y are actually from 1 to 0, so multiply your integral by -1 to reverse them. After doing that, W|A gives an answer here which is equivalent to the answer it gives for the original integral. Thanks from topsquark and Joppy
 April 23rd, 2017, 12:03 AM #6 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Thanks for your help

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