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     \lim_{a\to\infty} \displaystyle\int_{0}^{a} \dfrac{sinx}{x}dx

    It has a cool answer.
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    (Original post by Louisb19)
     \lim_{a\to\infty} \displaystyle\int_{0}^{a} \dfrac{sinx}{x}dx

    It has a cool answer.
    Off the top of my head, I can think of two ways to do this:

    DUTIS:
    Spoiler:
    Show

    \displaystyle I(\alpha) = \int_0^{\infty} \frac{\sin x}{x} e^{\alpha x} \, \mathrm{d}x.

    Then \displaystyle I'(\alpha) = \int_0^{\infty}\frac{\delta }{\delta \alpha} \left( \frac{\sin x}{x} e^{-\alpha x}\right) \, \mathrm{d}x = \int_0^{\infty} \sin x e^{-\alpha x} \, \mathrm{d}x =- \frac{1}{1+\alpha^2}

    Integrating: \displaystyle I(\alpha) = -\arctan \alpha + C, take \alpha = \infty to see that \displaystyle C = \frac{\pi}{2} and hence \displaystyle I(\alpha) = \frac{\pi}{2} - \arctan \alpha.

    Then our integral is \displaystyle I(0) = \frac{\pi}{2}

    Laplace transforms:
    Spoiler:
    Show
    \displaystyle \int_{0}^{\infty}\frac{\sin x}x \, \mathrm{d} x = L\{\sin x \}=\int_{0}^{\infty}\frac 1{s^{2}+1}\, \mathrm{d}s=\lim_{s \to \infty}\arctan s=\pi/2
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    (Original post by Zacken)
    Off the top of my head, I can think of two ways to do this:

    DUTIS:
    Spoiler:
    Show

    \displaystyle I(\alpha) = \int_0^{\infty} \frac{\sin x}{x} e^{\alpha x} \, \mathrm{d}x.

    Then \displaystyle I'(\alpha) = \int_0^{\infty}\frac{\delta }{\delta \alpha} \left( \frac{\sin x}{x} e^{-\alpha x}\right) \, \mathrm{d}x = \int_0^{\infty} \sin x e^{-\alpha x} \, \mathrm{d}x =- \frac{1}{1+\alpha^2}

    Integrating: \displaystyle I(\alpha) = -\arctan \alpha + C, take \alpha = \infty to see that \displaystyle C = \frac{\pi}{2} and hence \displaystyle I(\alpha) = \frac{\pi}{2} - \arctan \alpha.

    Then our integral is \displaystyle I(0) = \frac{\pi}{2}
    Laplace transforms:
    Spoiler:
    Show
    \displaystyle \int_{0}^{\infty}\frac{\sin x}x \, \mathrm{d} x = L\{\sin x \}=\int_{0}^{\infty}\frac 1{s^{2}+1}\, \mathrm{d}s=\lim_{s \to \infty}\arctan s=\pi/2
    It's a pretty cool answer isn't it.
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    (Original post by Louisb19)
    It's a pretty cool answer isn't it.
    Indeed, can you hazard a guess as to what \displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x} \, \mathrm{d}x is?
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    (Original post by Zacken)
    Indeed, can you hazard a guess as to what \displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x} \, \mathrm{d}x is?
    I'd guess since  f(x) = \dfrac{\sin x}{x} = f(-x) = \dfrac{\sin (-x)}{-x} (I think?)

    It would be  2\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} dx = \pi
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    (Original post by Louisb19)
    I'd guess since  f(x) = \dfrac{\sin x}{x} = f(-x) = \dfrac{\sin (-x)}{-x} (I think?)

    It would be  2\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} dx = \pi
    Precisely. \sin (-x) = -\sin x so f(x) is an even function.
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    So how about

    \displaystyle \int_{0}^{\infty} \frac{\sin^2 x}{x^2} \, \mathrm{d}x ?
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    (Original post by Zacken)
    Off the top of my head, I can think of two ways to do this:

    DUTIS:
    Spoiler:
    Show

    \displaystyle I(\alpha) = \int_0^{\infty} \frac{\sin x}{x} e^{\alpha x} \, \mathrm{d}x.

    Then \displaystyle I'(\alpha) = \int_0^{\infty}\frac{\delta }{\delta \alpha} \left( \frac{\sin x}{x} e^{-\alpha x}\right) \, \mathrm{d}x = \int_0^{\infty} \sin x e^{-\alpha x} \, \mathrm{d}x =- \frac{1}{1+\alpha^2}

    Integrating: \displaystyle I(\alpha) = -\arctan \alpha + C, take \alpha = \infty to see that \displaystyle C = \frac{\pi}{2} and hence \displaystyle I(\alpha) = \frac{\pi}{2} - \arctan \alpha.

    Then our integral is \displaystyle I(0) = \frac{\pi}{2}

    Laplace transforms:
    Spoiler:
    Show
    \displaystyle \int_{0}^{\infty}\frac{\sin x}x \, \mathrm{d} x = L\{\sin x \}=\int_{0}^{\infty}\frac 1{s^{2}+1}\, \mathrm{d}s=\lim_{s \to \infty}\arctan s=\pi/2
    the Laplace approach is not quite correct
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    (Original post by TeeEm)
    the Laplace approach is not quite correct
    Is this correct (sorry, my Laplace is horridly rusty):

    \displaystyle L\left\{\frac{\sin x}{x}\right\} = \int_0^{\infty} L\{\sin t\}\, \mathrm{d}s = \int_0^{\infty} \frac{1}{s^2 + 1} \, \mathrm{d}s = \frac{\pi}{2}
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    (Original post by Gregorius)
    So how about

    \displaystyle \int_{0}^{\infty} \frac{\sin^2 x}{x^2} \, \mathrm{d}x ?
    Is this just using
    Spoiler:
    Show
    parts with u = \sin^2 x and \mathrm{d}v = \dfrac{1}{x^2} with a quick x \mapsto 2x for the resulting integral?
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    (Original post by Louisb19)
     \lim_{a\to\infty} \displaystyle\int_{0}^{a} \dfrac{sinx}{x}dx

    It has a cool answer.
    You might like this. You come up with quite a few interesting problems, so you can post them there in the future!
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    (Original post by Zacken)
    Is this correct (sorry, my Laplace is horridly rusty):

    \displaystyle L\left\{\frac{\sin x}{x}\right\} = \int_0^{\infty} L\{\sin t\}\, \mathrm{d}s = \int_0^{\infty} \frac{1}{s^2 + 1} \, \mathrm{d}s = \frac{\pi}{2}
    I just got back ...
    food first!
    Then I will write and post the method
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    (Original post by TeeEm)
    I just got back ...
    food first!
    Then I will write and post the method
    Bon appetit!
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    (Original post by Zacken)
    Bon appetit!
    here is the one and another 3 for the interest of whoever cares

    Name:  1.jpg
Views: 69
Size:  302.1 KBAttachment 475265475267Attachment 475265475267475260[attach]4.752654752674753e+23[/attach]


    I posted earlier another integral problem here
    http://www.thestudentroom.co.uk/show....php?t=3327325

    Maybe this is easy to be done now
    Attached Images
       
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    (Original post by TeeEm)
    here is the one and another 3 for the interest of whoever cares

    Name:  1.jpg
Views: 69
Size:  302.1 KBAttachment 475265475267Attachment 475265475267475260[attach]4.752654752674753e+23[/attach]


    I posted earlier another integral problem here
    http://www.thestudentroom.co.uk/show....php?t=3327325

    Maybe this is easy to be done now
    Many thanks!
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    (Original post by Zacken)
    Is this just using
    Spoiler:
    Show
    parts with u = \sin^2 x and \mathrm{d}v = \dfrac{1}{x^2} with a quick x \mapsto 2x for the resulting integral?
    Looks about right; what is amusing in the context of the original question is the actual value of the answer...

    Unfortunately the pattern does not continue for

     \displaystyle \int_{0}^{\infty} \frac{\sin^n x}{x^n} \, \mathrm{d}x

    for n in general...
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    (Original post by Gregorius)
    Looks about right; what is amusing in the context of the original question is the actual value of the answer...Unfortunately the pattern does not continue for \displaystyle \int_{0}^{\infty} \frac{\sin^n x}{x^n} \, \mathrm{d}xfor n in general...
    I was just about to ask about that. :lol:

    Interestingly enough we do have that becomes \frac{3\pi}{8}
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    (Original post by Gregorius)
    Looks about right; what is amusing in the context of the original question is the actual value of the answer...

    Unfortunately the pattern does not continue for

     \displaystyle \int_{0}^{\infty} \frac{\sin^n x}{x^n} \, \mathrm{d}x

    for n in general...
    I'll be back in half an hour or so, I'm going to play around with that integral and see if I can any nice recurrences or patterns! Thanks for this!
 
 
 
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