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April 20th, 2017, 12:19 PM   #1
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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.
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April 20th, 2017, 02:39 PM   #2
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If you substitute y = 1/x, what are the values of y that correspond to the given limits for x?
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April 22nd, 2017, 02:58 AM   #3
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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.
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April 22nd, 2017, 08:15 AM   #4
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This is a Fresnel integral

How to solve
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April 22nd, 2017, 08:52 AM   #5
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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.
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April 23rd, 2017, 12:03 AM   #6
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Thanks for your help
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