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Thread: An arctan integral

  1. #1
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    An arctan integral

    Evaluate $\displaystyle \int_{0}^{1}{\frac{\arctan \left( \sqrt{2+x^{2}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}.$
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  2. #2
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    okay the problem at first glance it's hard, but i don't want to give any hint yet, so i'll wait a couple of days and i'll post a solution.
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  3. #3
    Super Member Deadstar's Avatar
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    Posting in thread so I remember about it. I think it has to do with...
    $\displaystyle \frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$.

    Then find an expression for arctan(x)/x maybe and use integration by parts. I'll look at it later.
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  4. #4
    Super Member Deadstar's Avatar
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    Will just post any working as I go along...

    Spoiler:

    Let $\displaystyle y = \sqrt{2 + x^2}$, $\displaystyle dy = \frac{x}{\sqrt{2 + x^2}}
    dx$

    Then for limits we have $\displaystyle 0 \to \sqrt{2}$ and $\displaystyle 1 \to \sqrt{3}$

    So our new integral is $\displaystyle \int_{\sqrt{2}}^{\sqrt{3}} \frac{\arctan(y)}{(y^2 - 1)(\sqrt{y^2 - 2})} dy$
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  5. #5
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    I make use of (again) 'magic differentiation' to solve this problem .


    Consider $\displaystyle \int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} } $

    Sub. $\displaystyle x = \frac{1}{t} $ , it becomes

    $\displaystyle \int_1^{\infty} \frac{t~dt}{(t^2+1)\sqrt{1+2t^2} }$

    sub $\displaystyle 1+2t^2 = u^2 $ .... After a few steps , we can obtain

    $\displaystyle \int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} } = \frac{\pi}{6}$


    The integral


    $\displaystyle \int_{0}^{1}{\frac{\arctan \left( \sqrt{2+x^{2}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$

    $\displaystyle = \int_{0}^{1}{\frac{\frac{\pi}{2} - \arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$

    $\displaystyle = \frac{\pi}{2} \int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} } - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$


    $\displaystyle = \frac{\pi^2}{12} - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$


    Consider $\displaystyle I(a) = \int_{0}^{1}{\frac{\arctan \left( \frac{a}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx} $

    Magic Differentiation !

    $\displaystyle I'(a) = \int_0^1 \frac{dx}{(x^2+1)(x^2 + a^2 + 2 )} $

    $\displaystyle = \frac{1}{1 + a^2 } \int_0^1 \left[ \frac{1}{1 + x^2 } - \frac{1}{x^2 + a^2+2} \right]~dx $

    $\displaystyle = \frac{\pi}{4}\cdot \frac{1}{1+a^2} - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+a^{2}}} \right)}{\left( 1+a^{2} \right)\sqrt{2+a^{2}}}\,dx}$

    So $\displaystyle \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx} = I(1) = I(1)-I(0)$

    $\displaystyle = \frac{\pi^2}{16}- I(1) $

    $\displaystyle = \frac{\pi^2}{32} $


    The answer to the problem is $\displaystyle \frac{\pi^2}{12} - \frac{\pi^2}{32} $

    $\displaystyle = \frac{5 \pi^2}{96} $
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  6. #6
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    okay, my solution uses the fact that

    $\displaystyle \frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\sqrt{2+x^{2}}}=\int_{0}^{1}{\frac{dt}{2+ x^{2}+t^{2}}}.$
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  7. #7
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Krizalid View Post
    okay, my solution uses the fact that

    $\displaystyle \frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\sqrt{2+x^{2}}}=\int_{0}^{1}{\frac{dt}{2+ x^{2}+t^{2}}}.$
    Did you write the original integral as a triple integral and then change the order of integration?
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  8. #8
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    as a double one.
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  9. #9
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Krizalid View Post
    as a double one.
    $\displaystyle \int^{1}_{0} \int^{\infty}_{1} \frac{dt \ dx}{(1+x^{2})(2+x^{2}+t^{2})} $ ?
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  10. #10
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    sorry for the late answer, actually, it holds for a better way.

    try to find it, my time is bounded by now.
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