MHB Fresnel Integrals: Unsolved Question from MHF

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    Fresnel Integrals
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The discussion centers on an unsolved question regarding the integral of sin(t^2) from 0 to infinity, specifically proving that it equals sqrt(pi/8). Participants are encouraged to share hints or alternative methods for solving this integral, with contour integration being one suggested approach. A reference to a Spanish PDF is provided, although a typo in the document is noted regarding the integral's definition. Another method mentioned is the Laplace transform, although details are not included in the original forum post. The conversation invites further exploration of various mathematical techniques to tackle this integral challenge.
Fernando Revilla
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I quote a unsolved question posted in MHF by user poorbutttryagin on February 5th, 2013.
I read 'functions of one complex variable by Conway'

186pg, 7.7. Prove that int_0^inf sin(t^2) dt = sqrt(pi/8)

What is the starting point?

Any comment or hint is welcomed !

Thanks !
 
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Have a look at the pdf here:

http://www.fernandorevilla.es/iii/paginas-111-120/120-integrales-de-fresnel

P.S. 1 Although it is in Spanish, I think that one can follow the outline looking only at the formulas.

P.S. 2 There is a typo in the second line of the pdf.:

It should be $I_2=\displaystyle\int_0^{+\infty}\sin x^2\;dx$ instead of $I_2=\displaystyle\int_0^{+\infty}\cos x^2\;dx$
 
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I see that you are using contour integration to solve the integral .

Do you have another method to solve it ?
 
ZaidAlyafey said:
Do you have another method to solve it ?

I know another metod (Laplace transform), but it is not in my site. Have a look (for example) here:

http://www.mymathforum.com/viewtopic.php?f=22&t=20045
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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