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Background: This is another proof that the fresnel integrals $C(\infty)$ and $S(\infty)$ are 1/2 from Hobson, Riley, Bence, Mathematical Methods chapter 25.8 the stationary phase method.

enter image description here According to here, the integral leading to the underlined does not equal 0 but diverges. According to here the obvious but wrong answer that it equals 0 for being odd is right if you say it's the Cauchy principal value of an ill defined integral. However in the above proof, they evaluated the odd integral to equal twice the integral from zero. Is there any explanation for why and could I do this to any improper infinite integral? $$\int_{-\infty}^{\infty}\sin(\frac{1}{2}\pi u^2)\text{du}=2\int_0^{\infty}\sin(\frac{1}{2}\pi u^2)\text{du}$$

user5389726598465
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    Because you're integrating an even function? – Angina Seng May 13 '17 at 07:30
  • I'm talking about the sin(). Does that switch from odd to even at infinity? – user5389726598465 May 13 '17 at 07:31
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    Both links are not relevant here because you have to integrate $\sin u^2$ and not $\sin u$ – gammatester May 13 '17 at 07:42
  • I think it would be helpful here to plot the graph of $\sin (\frac{1}{2} \pi u^2)$ to see that it really is symmetric around 0. Then it only needs to be seen that the integral converges. This can be justified heuristically by noting that the area under the $n^{th}$ peak scales as the inverse square of $n$ – Mark May 13 '17 at 07:47
  • I forgot about the $u^2$. @Lord thanks for not telling me that's what you meant like 30 minutes ago...! – user5389726598465 May 13 '17 at 08:05

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Here is a plot of $\sin(u^2 \pi/2)$. The integral over the real line, if it exists, must be double the integral over the positive real line because of symmetry. sin(u^2 \pi /2)

The fact that the integral exists on the positive real line, as you mentioned, can be proven by appealing to existence of Fresnel integral.

Mark
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