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Thread: Fun integral

  1. #1
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    Fun integral

    Show that:

    $\displaystyle \int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)} dx$

    is independent of $\displaystyle \alpha$.

    This took me quite a while to figure out.
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  2. #2
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    and took me 1 second to tell ya that it's a quite known problem!

    so leave others to try it! (i'm telling you people who already knows the answer.)
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  3. #3
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    Did anyone else do a facepalm when they finally figured it out? I stared blankly at my paper for 15 minutes trying to figure it out, and I feel dumb now for not seeing it much sooner.

    I was never good at integrals to begin with, however .
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  4. #4
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    Quote Originally Posted by hjortur View Post
    Show that:

    $\displaystyle \int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)} dx$

    is independent of $\displaystyle \alpha$.

    This took me quite a while to figure out.
    It is also a constant when $\displaystyle \alpha = 0 $
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    How is this integral done? I'm thinking complex contour integration, but I bet there's an easier way.

    Also, I've noticed that this works for all $\displaystyle \alpha\in\mathbb{C} $.
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  6. #6
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    use $\displaystyle \tan x=\frac{\sin x}{\cos x}$ and substitute $\displaystyle x\mapsto \frac\pi2-x.$
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Krizalid View Post
    use $\displaystyle \tan x=\frac{\sin x}{\cos x}$ and substitute $\displaystyle x\mapsto \frac\pi2-x.$
    After doing that, I get $\displaystyle \int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{\alpha}(x) } = \int_0^{\frac{\pi}{2}}\frac{dx}{1+\cot^{\alpha}(x) } = \int_0^{\infty}\frac{du}{u^2+u^{2-\alpha}+u^{-\alpha}+1} $.

    I'm definitely missing something here...
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  8. #8
    Super Member Random Variable's Avatar
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    I think it's the following:

    $\displaystyle \int_{0}^{\pi /2} \frac{1}{1+\tan^{\alpha}x} \ dx = \int^{\pi /2}_{0}\frac{\cos^{\alpha}x}{\cos^{\alpha}x + \sin^{\alpha} x} \ dx $

    Let $\displaystyle x = \frac{\pi}{2}-x $

    $\displaystyle = -\int^{0}_{ \pi /2}\frac{\cos^{\alpha}(\pi/ 2-x)}{\cos^{\alpha}(\pi / 2-x) + \sin^{\alpha} (\pi / 2 -x)} \ dx = \int^{\pi /2}_{0}\frac{\sin^{\alpha}x}{\sin^{\alpha}x + \cos^{\alpha} x} \ dx $

    so $\displaystyle \int_{0}^{\pi /2} \frac{1}{1+\tan^{\alpha}x} \ dx = \frac{1}{2}\int^{\pi /2}_{0}\frac{\cos^{\alpha}x + \sin^{\alpha} x}{\cos^{\alpha}x + \sin^{\alpha} x} \ dx = \frac{1}{2}\int^{\pi /2}_{0} dx = \frac{\pi}{4}$
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  9. #9
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