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Math Help - improper arctangent integral

  1. #1
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    Post improper arctangent integral

    ?
    send me the full solution to solve this!!!!!!!!
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  2. #2
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    Quote Originally Posted by kamaksh_ice View Post
    ∫_0^∞((tan^(-1)ax-tan^(-1)bx) )/ax dx
    Fix it!
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    reply me de full solution
    or else what?

    Quote Originally Posted by kamaksh_ice View Post
    ∫_0^∞((tan^(-1)ax-tan^(-1)bx) )/ax dx
    what are these ? signs about?

    did you mean: \int_{0}^{\infty} \frac {\tan^{-1} ax - \tan^{-1} bx}{ax}~dx ?
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  4. #4
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    Amazingly this integral appears in my Advanced Calculus book so the credit is not mine (though I wish it was).

    First you must say 0<b<a. Confirm it is uniformly continous and do the following double integral trick.

    \int_0^{\infty} \frac{\tan^{-1} ax - \tan^{-1} bx}{x} dx = \int_0^{\infty} \int_b^a \frac{1}{t^2 x^2 + 1} \ dt \ dx.
    Change order,
    \int_b^a \int_0^{\infty} \frac{1}{t^2x^2+1} \ dx \ dt = \int_b^a \frac{\tan^{-1} tx}{t} \big|_0^{\infty} \ dx = \int_a^b \frac{\pi}{2t} dt = \frac{\pi}{2} \ln \frac{a}{b}
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  5. #5
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    More generally

    If f'(x) is continuous and the integral converges,

    \int_0^\infty\frac{f(ax)-f(bx)}x\,dx=(f(0)-f(\infty))\ln\frac ba

    This beautiful result is called Frullani's Integral.

    Nice, isn't it?
    Last edited by Krizalid; September 24th 2007 at 06:27 PM.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    More generally

    If f'(x) is continuous and the integral converges,

    \int_0^\infty\frac{f(ax)-f(bx)}x\,dx=(f(0)-f(\infty))\ln\frac ab

    This beautiful result is called Frullani's Integral.

    Nice, isn't it?
    Yes it is nice . Where did you get that from? It seems MathWorld has it while Wikipedia does not.

    You should also state the conditions on the function* and limits. Perhaps a>b>0.


    *)Maybe it should also be uniformly continous and having a limit at infinity?
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  7. #7
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    I found (not mine) a proof of Frullani's Integral, and I'd like to share it

    Let

    \int_\varphi ^\tau {\frac{{f(ax) - f(bx)}}<br />
{x}\,dx}.

    So

    \int_\varphi ^\tau {\frac{{f(ax) - f(bx)}}<br />
{x}\,dx} = \int_\varphi ^\tau {\frac{{f(ax)}}<br />
{x}\,dx} - \int_\varphi ^\tau {\frac{{f(bx)}}<br />
{x}\,dx} = \int_{a\varphi }^{a\tau } {\frac{{f(x)}}<br />
{x}\,dx} - \int_{b\varphi }^{b\tau } {\frac{{f(x)}}<br />
{x}\,dx} .

    And

    \int_\varphi ^\tau {\frac{{f(ax) - f(bx)}}<br />
{x}\,dx} = \int_{a\varphi }^{b\varphi } {\frac{{f(x)}}<br />
{x}\,dx} - \int_{a\tau }^{b\tau } {\frac{{f(x)}}<br />
{x}\,dx} = \int_a^b {\frac{{f(u\varphi )}}<br />
{u}\,du} - \int_a^b {\frac{{f(u\tau )}}<br />
{u}\,du} .

    Now let \varphi \to 0,\,\tau \to \infty ,

    \int_a^b {\frac{{f(0)}}<br />
{u}\,du} = f(0)\ln \frac{b}<br />
{a} & \int_a^b {\frac{{f(\infty )}}<br />
{u}\,du} = f(\infty )\ln \frac{b}<br />
{a}.

    The conclusion follows.
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