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  1. #1
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    Difficult integral

    How can I calculate the integral of sin(x^2) from zero to infinity? Do I need to use the residue theorem?
    I need a detailed answer, because i'm very new in this subject.
    Thanks for any kind of help
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  2. #2
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    Quote Originally Posted by Gina88 View Post
    How can I calculate the integral of sin(x^2) from zero to infinity? Do I need to use the residue theorem?
    I need a detailed answer, because i'm very new in this subject.
    Thanks for any kind of help
    View Wikipedia Fresnel integral
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  3. #3
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    I'm sorry, but I didn't understand.
    I know I can write sin(x^2) like this:x^2 - x^6/3! + x^10/5! - x^14/7!...
    And then what?
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  4. #4
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    Quote Originally Posted by Gina88 View Post
    I'm sorry, but I didn't understand.
    I know I can write sin(x^2) like this:x^2 - x^6/3! + x^10/5! - x^14/7!...
    And then what?
    Integrate e^{iz^2} over the contour C shown in the link you were given. Note that by Cauchy's theorem \int_C e^{iz^2} \, dz = 0.
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  5. #5
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    start with the well-known and easy to prove intgeral \int_0^{\infty}e^{-x^2} \ dx = \frac{\sqrt{\pi}}{2}. let x^2=st, where we choose s to be the variable. then we'll get \frac{1}{\sqrt{t}}=\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{s}}e^{-st} \ ds. \ \ \ \ \ \ (1)

    also it's easy to see that \int_0^{\infty} e^{-ax} \sin x \ dx = \frac{1}{a^2+1}. \ \ \ \ \ \ (2)

    let J=\int_0^{\infty} \frac{dx}{\sqrt{x}(x^2+1)} and x=z^2. then J=2 \int_0^{\infty} \frac{dz}{z^4 + 1}. now let u=z - \frac{1}{z}, \ \ v = z+\frac{1}{z}. then: 2\int \frac{dz}{z^4 + 1}=\int \frac{du}{u^2 + 2} + \int \frac{dv}{v^2-2}=\frac{1}{\sqrt{2}} \tan^{-1}(u/\sqrt{2}) + \frac{1}{2\sqrt{2}} \ln \left|\frac{v-\sqrt{2}}{v+\sqrt{2}} \right|,

    which gives us: J=\frac{\pi}{\sqrt{2}}. \ \ \ \ \ \ \ (3)

    now let I=\int_0^{\infty} \sin x^2 \ dx and put x^2=t. then by (1): I=\frac{1}{2} \int_0^{\infty} \frac{\sin t}{\sqrt{t}} \ dt = \frac{1}{2 \sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{s}} \int_0^{\infty} e^{-st} \sin t \ dt \ ds. so by (2) and (3): I=\frac{1}{2 \sqrt{\pi}} \int_0^{\infty}\frac{ds}{\sqrt{s}(s^2+1)}=\frac{1}  {2\sqrt{\pi}} \cdot \frac{\pi}{\sqrt{2}}=\sqrt{\frac{\pi}{8}}.
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