Όλοκλήρωμα τύπου Fresnel

galactus · #1

Hi Everyone. It's been a while. Nice to be back.

Sorry for the English. Maybe someone may be kind enough to translate.

But, I have a challenging integral to post you may like.

Show that:

$\displaystyle \int_{0}^{\infty}\frac{\sin(x^{2})}{\pi-x^{2}}dx=\frac{\sqrt{\pi}}{2}\left(C(\sqrt{2})+S(\sqrt{2})\right)$

where C and S are the Fresnel integrals defined as:

$\displaystyle C(u)=\int_{0}^{u}\cos(\frac{\pi}{2}x^{2})dx, \;\ S(u)=\int_{0}^{u}\sin(\frac{\pi}{2}x^{2})dx$

In this case, $u=\sqrt{2}$

#2

galactus έγραψε:Hi Everyone. It's been a while. Nice to be back.

Sorry for the English. Maybe someone may be kind enough to translate.

But, I have a challenging integral to post you may like.

Show that:

$\displaystyle \int_{0}^{\infty}\frac{\sin(x^{2})}{\pi-x^{2}}dx=\frac{\sqrt{\pi}}{2}\left(C(\sqrt{2})+S(\sqrt{2})\right)$

where C and S are the Fresnel integrals defined as:

$\displaystyle C(u)=\int_{0}^{u}\cos(\frac{\pi}{2}x^{2})dx, \;\ S(u)=\int_{0}^{u}\sin(\frac{\pi}{2}x^{2})dx$

In this case, $u=\sqrt{2}$

Μετάφραση: Γειά σας. Έχει περάσει κάποιος καιρός. Χαίρομαι που επέστρεψα. Συγγνώμη για τα Αγγλικά. Ίσως κάποιος έχει την καλοσύνη να το μεταφράσει.

Έχω ένα ενδιαφέρον ολοκλήρωμα να προτείνω, που μπορεί να σας αρέσει.

Δείξτε ότι

$\displaystyle \int_{0}^{\infty}\frac{\sin(x^{2})}{\pi-x^{2}}dx=\frac{\sqrt{\pi}}{2}\left(C(\sqrt{2})+S(\sqrt{2})\right)$

όπου C και S είναι ταολοκληρώματα Fresnel που ορίζονται ως:

$\displaystyle C(u)=\int_{0}^{u}\cos(\frac{\pi}{2}x^{2})dx, \;\ S(u)=\int_{0}^{u}\sin(\frac{\pi}{2}x^{2})dx$

Σε αυτήν την περίπτωση $u=\sqrt{2}$

#3

galactus έγραψε:Show that: $\displaystyle \int_{0}^{\infty}\frac{\sin(x^{2})}{\pi-x^{2}}dx=\frac{\sqrt{\pi}}{2}\left(C(\sqrt{2})+S(\sqrt{2})\right)$

Λήμμα $\displaystyle{L.1:\int\limits_0^\infty {\frac{{\cos ax}}{{\sqrt x }}dx} = \frac{1}{2}\int\limits_0^\infty {\frac{{{e^{iax}} + {e^{ - iax}}}}{{\sqrt x }}dx} = \frac{{\sqrt \pi }}{2}\left( {\frac{1}{{\sqrt { - ia} }} + \frac{1}{{\sqrt {ia} }}} \right) = \frac{{\sqrt \pi }}{{2\sqrt a }}\left( {\frac{{\sqrt 2 }}{{1 - i}} + \frac{{\sqrt 2 }}{{1 + i}}} \right) = \sqrt {\frac{\pi }{{2a}}} }$

Λήμμα $\displaystyle{L.2:\int\limits_0^\infty {\frac{{\sin ax}}{{\sqrt x }}dx} = \sqrt {\frac{\pi }{{2a}}} }$ (όμοια)

Λήμμα $\displaystyle{L.3:\frac{d}{{da}}\left( {C\left( {\sqrt {\frac{{2a}}{\pi }} } \right)} \right) = \frac{d}{{da}}\left( {\int\limits_0^{\sqrt {2a/\pi } } {\cos \left( {\frac{\pi }{2}{x^2}} \right)dx} } \right) = \frac{1}{{\sqrt {2a\pi } }}\cos a}$ και $\displaystyle{\frac{d}{{da}}\left( {S\left( {\sqrt {\frac{{2a}}{\pi }} } \right)} \right) = \frac{1}{{\sqrt {2a\pi } }}\sin a}$ (στοιχειώδη).
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$\displaystyle{\int\limits_0^\infty {\frac{{\sin {x^2}}}{{\pi - {x^2}}}dx} = \mathop = \limits^{x \to \sqrt \pi \cdot x} = \frac{1}{{\sqrt \pi }}\int\limits_0^\infty {\frac{{\sin \pi {x^2}}}{{1 - {x^2}}}dx} = \frac{1}{{\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\sin \pi {x^2}}}{{1 - {x^2}}}dx} - \int\limits_1^\infty {\frac{{\sin \pi {x^2}}}{{{x^2} - 1}}dx} } \right) = \mathop = \limits^{{x^2} \to \sqrt x } = }$

$\displaystyle{ = \frac{1}{{2\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\sin \pi x}}{{\sqrt x \left( {1 - x} \right)}}dx} - \int\limits_1^\infty {\frac{{\sin \pi y}}{{\sqrt y \left( {y - 1} \right)}}dy} } \right) = \mathop = \limits_{y \to 1 + y}^{x \to 1 - x} = \frac{1}{{2\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\sin \pi x}}{{x\sqrt {1 - x} }}dx} + \int\limits_1^\infty {\frac{{\sin \pi y}}{{y\sqrt {1 + y} }}dy} } \right)}$

Έστω $\displaystyle{f\left( a \right) = \frac{1}{{2\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\sin ax}}{{x\sqrt {1 - x} }}dx} + \int\limits_1^\infty {\frac{{\sin ay}}{{y\sqrt {1 + y} }}dy} } \right)}$ . Τότε

$\displaystyle{f'\left( a \right) = \frac{1}{{2\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\cos ax}}{{\sqrt {1 - x} }}dx} + \int\limits_1^\infty {\frac{{\cos ay}}{{\sqrt {1 + y} }}dy} } \right) = \mathop = \limits_{1 + y \to y}^{1 - x \to x} = \frac{1}{{2\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\cos \left( {a - ax} \right)}}{{\sqrt x }}dx} } \right. + }$

$\displaystyle{\left. { + \int\limits_1^\infty {\frac{{\cos \left( {ay - y} \right)}}{{\sqrt y }}dy} } \right) = \frac{1}{{2\sqrt \pi }}\int\limits_0^\infty {\frac{{\cos \left( {a - ax} \right)}}{{\sqrt x }}dx} = \frac{1}{{2\sqrt \pi }}\left( {\cos a \cdot \int\limits_0^\infty {\frac{{\cos ax}}{{\sqrt x }}dx} + } \right.}$

$\displaystyle{\left. { + \sin a \cdot \int\limits_0^\infty {\frac{{\sin ax}}{{\sqrt x }}dx} } \right) = \mathop = \limits^{L.1.2} = \frac{1}{{2\sqrt {2a} }}\left( {\cos a + \sin a} \right) = \frac{{\sqrt \pi }}{2}\left( {\frac{1}{{\sqrt {2a\pi } }}\cos a + \frac{1}{{\sqrt {2a\pi } }}\sin a} \right) = }$

$\displaystyle{ = \mathop = \limits^{L.3} = \frac{{\sqrt \pi }}{2}\frac{d}{{da}}\left( {S\left( {\sqrt {\frac{{2a}}{\pi }} } \right) + C\left( {\sqrt {\frac{{2a}}{\pi }} } \right)} \right)}$

Οπότε $\displaystyle{f\left( a \right) = \frac{{\sqrt \pi }}{2}\left( {S\left( {\sqrt {\frac{{2a}}{\pi }} } \right) + C\left( {\sqrt {\frac{{2a}}{\pi }} } \right)} \right) + c}$ . Για $\displaystyle{a = 0 \Rightarrow c = 0}$ . Άρα $\displaystyle{f\left( a \right) = \frac{{\sqrt \pi }}{2}\left( {S\left( {\sqrt {\frac{{2a}}{\pi }} } \right) + C\left( {\sqrt {\frac{{2a}}{\pi }} } \right)} \right)}$ .

και για $\displaystyle{a = \pi }$ προκύπτει $\displaystyle{\int\limits_0^\infty {\frac{{\sin {x^2}}}{{\pi - {x^2}}}dx} = \frac{1}{{2\sqrt \pi }}\left( {\int\limits_0^1 {\frac{{\sin \pi x}}{{x\sqrt {1 - x} }}dx} + \int\limits_1^\infty {\frac{{\sin \pi y}}{{y\sqrt {1 + y} }}dy} } \right) = f\left( \pi \right) = \frac{{\sqrt \pi }}{2}\left( {S\left( {\sqrt 2 } \right) + C\left( {\sqrt 2 } \right)} \right)}$

Ένα ενδιαφέρον ενδιάμεσο αποτέλεσμα είναι ότι $\displaystyle{\int\limits_0^1 {\frac{{\sin \pi {x^2}}}{{1 - {x^2}}}} dx = \frac{\pi }{2}\left( {{S^2}\left( {\sqrt 2 } \right) + {C^2}\left( {\sqrt 2 } \right)} \right)}$

Όλοκλήρωμα τύπου Fresnel

Όλοκλήρωμα τύπου Fresnel

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Re: Όλοκλήρωμα τύπου Fresnel

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