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Math Help - Another integral

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Another integral

    Ok how would you go about doing this one \int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}dx?....I would imagine you go \int_0^{\infty}\bigg[\sum_{n=0}^{\infty}\frac{(-1)^n\cdot{a^{2n+1}}\cdot{x^{2n}}}{2n+1}-\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{b^{2n+1}}\cdot{x^{2n}}}{2n+1}\bigg]{dx} =\bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{a^{2n+1}}\cdot{x^{2n+1}}}{(2n+1)^2}-\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{b^{2n+1}}\cdot{x^{2n+1}}}{(2n+1)^2}\bi  gg]\bigg|_0^{\infty}
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    Quote Originally Posted by Mathstud28 View Post
    Ok how would you go about doing this one \int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}dx?....I would imagine you go \int_0^{\infty}\bigg[\sum_{n=0}^{\infty}\frac{(-1)^n\cdot{a^{2n+1}}\cdot{x^{2n}}}{2n+1}-\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{b^{2n+1}}\cdot{x^{2n}}}{2n+1}\bigg]{dx} =\bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{a^{2n+1}}\cdot{x^{2n+1}}}{(2n+1)^2}-\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{b^{2n+1}}\cdot{x^{2n+1}}}{(2n+1)^2}\bi  gg]\bigg|_0^{\infty}
    There are many possible approaches. Here's a simple one:

    Let f(a) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx.


    Then f'(a) = \frac{d}{da} \left[\int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx\right] = \int_{0}^{\infty} \frac{d}{da} \left[ \frac{\tan^{-1}(ax)}{x} \right] \, dx


     = \int_{0}^{\infty} \frac{1}{a^2 x^2 + 1} \, dx = \frac{1}{a} \, \tan^{-1}(ax) \bigg{|}_{0}^{\infty} = \frac{\pi}{2a}.


    Therefore f(a) = \int \frac{\pi}{2a} \, da = \frac{\pi}{2} \ln (a) + C.


    Therefore f(a) - f(b) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx - \int_{0}^{\infty} \frac{\tan^{-1}(bx)}{x} \, dx


     = \left(\frac{\pi}{2} \ln (a) + C\right) - \left(\frac{\pi}{2} \ln (b) + C\right) = \frac{\pi}{2} \ln\left( \frac{a}{b}\right).


    I'm sure the big K will outline the other approaches
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    There are many possible approaches. Here's a simple one:

    Let f(a) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx.


    Then f'(a) = \frac{d}{da} \left[\int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx\right] = \int_{0}^{\infty} \frac{d}{da} \left[ \frac{\tan^{-1}(ax)}{x} \right] \, dx


     = \int_{0}^{\infty} \frac{1}{a^2 x^2 + 1} \, dx = \frac{1}{a} \, \tan^{-1}(ax) \bigg{|}_{0}^{\infty} = \frac{\pi}{2a}.


    Therefore f(a) = \int \frac{\pi}{2a} \, da = \frac{\pi}{2} \ln (a) + C.


    Therefore f(a) - f(b) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx - \int_{0}^{\infty} \frac{\tan^{-1}(bx)}{x} \, dx


     = \left(\frac{\pi}{2} \ln (a) + C\right) - \left(\frac{\pi}{2} \ln (b) + C\right) = \frac{\pi}{2} \ln\left( \frac{a}{b}\right).
    That is a really neat method...but does that always work or does it work conveniently here...or could I always go \int_a^{b}F(x)dx=\int[f(b)-f(a)]\bigg|_a^{b}dx?
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    Quote Originally Posted by Mathstud28 View Post
    That is a really neat method...but does that always work or does it work conveniently here...or could I always go \int_a^{b}F(x)dx=\int[f(b)-f(a)]\bigg|_a^{b}dx?
    Read this (scroll down to examples): Differentiation under the integral sign - Wikipedia, the free encyclopedia

    The technique was used extensively by Feynman (you can read about that in this book: Amazon.com: Surely You're Joking, Mr. Feynman! (Adventures of a Curious Character): Richard P. Feynman,Ralph Leighton,Edward Hutchings,Albert R. Hibbs: Books)

    It can always be used with definite integrals, but will not always be useful. Obviously it was useful in this case.
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    Quote Originally Posted by mr fantastic View Post
    Read this (scroll down to examples): Differentiation under the integral sign - Wikipedia, the free encyclopedia

    The technique was used extensively by Feynman (you can read about that in this book: Amazon.com: Surely You're Joking, Mr. Feynman! (Adventures of a Curious Character): Richard P. Feynman,Ralph Leighton,Edward Hutchings,Albert R. Hibbs: Books)

    It can always be used with definite integrals, but will not always be useful. Obviously it was useful in this case.
    Thanks a lot!
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    \frac{{\arctan (\alpha x) - \arctan (\beta x)}}<br />
{x} = \int_\beta ^\alpha  {\frac{1}<br />
{{1 + x^2 y^2 }}\,dy}. So,

    \int_0^\infty  {\frac{{\arctan (\alpha x) - \arctan (\beta x)}}<br />
{x}\,dx}  = \int_0^\infty \! {\int_\beta ^\alpha  {\frac{{dy\,dx}}<br />
{{1 + x^2 y^2 }}} }. After reverse integration order and some straightforward calculations yields the answer.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    \frac{{\arctan (\alpha x) - \arctan (\beta x)}}<br />
{x} = \int_\beta ^\alpha {\frac{1}<br />
{{1 + x^2 y^2 }}\,dy}. So,

    \int_0^\infty {\frac{{\arctan (\alpha x) - \arctan (\beta x)}}<br />
{x}\,dx} = \int_0^\infty \! {\int_\beta ^\alpha {\frac{{dy\,dx}}<br />
{{1 + x^2 y^2 }}} }. After reverse integration order and some straightforward calculations yields the answer.
    Ok I see that
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  8. #8
    Moo
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    This is just because I can't remember, but what are the conditions for reversing the order of integration ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    This is just because I can't remember, but what are the conditions for reversing the order of integration ?
    easiest way is to graph the region of integration and change the functions of x to functions of y and vice versa. this will give you the new limits to change the integration order
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    Moo
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    I mean aren't there any condition on the function, such as C^1, derivability or whatever ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    I mean aren't there any condition on the function, such as C^1, derivability or whatever ?
    oh, well, in that case, i don't remember either . As far as reversing integration order goes, I think we worry about integrability, not differentiability. we would worry about differentiability when introducing the second integral in the first place.
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  12. #12
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    Quote Originally Posted by Moo View Post
    This is just because I can't remember, but what are the conditions for reversing the order of integration ?
    Scroll down to theorem 9.4 (Fubini's Theorem - Stronger Form): Repeated Integrals and Fubini's Theorem
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  13. #13
    Moo
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    Ok, so only continuity !
    Gosh... I've studied this theorem two weeks ago only :/

    Thanks
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