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    I have an answer involving f(\infty) and I assume the condition on f means that this is 0, but I don't see how.

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    What I've done so far:
    Write \int_{0}^{\infty }\frac{f(ax)-f(bx)}{x}dx = \int_{0}^{\infty }\int_{b}^{a}f'(xy)dydx
    Reversing order of integration gives \int_{b}^{a }\int_{0}^{\infty}f'(xy)dxdy =\int_{b}^{a }\left[ \frac{f(xy)}{y} \right]_{0}^{\infty}dy =\int_{b}^{a }(f(\infty)-f(0))\frac{1}{y}dy=(f(0)-f(\infty))log(\frac{b}{a})
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    (Original post by solC)
    I have an answer involving f(\infty) and I assume the condition on f means that this is 0, but I don't see how.
    Can't help with the general question.

    But i notice that if we let f(x)=sin(x), then this meets the criteria of the question, and f(\infty) doesn't exist, let alone equal zero.
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    Replacing \infty by N (in other words, using your argument, but with a finite upper limit that we will later let go to infinity) , you have \displaystyle \int_a^b \left[\dfrac{f(xy)}{y}\right]_0^N \, dy = \displaystyle \int_a^b [f(Ny)-f(0)]\frac{1}{y} \, dy

    But \displaystyle \int_a^b f(Ny)/y\, dy =  \int_{Na}^{Nb} f(x) \frac{1}{x}\,dx (*) (by the linear sub x = Ny).

    The given condition on \int f(x)\frac{1}{x}\,dx existing is enough to show that (*) goes to 0 as N goes to infinity.
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    (Original post by solC)
    Name:  VC.JPG
Views: 23
Size:  17.2 KB

    I have an answer involving f(\infty) and I assume the condition on f means that this is 0, but I don't see how.

    Spoiler:
    Show



    What I've done so far:
    Write \int_{0}^{\infty }\frac{f(ax)-f(bx)}{x}dx = \int_{0}^{\infty }\int_{b}^{a}f'(xy)dydx
    Reversing order of integration gives \int_{b}^{a }\int_{0}^{\infty}f'(xy)dxdy =\int_{b}^{a }\left[ \frac{f(xy)}{y} \right]_{0}^{\infty}dy =\int_{b}^{a }(f(\infty)-f(0))\frac{1}{y}dy=(f(0)-f(\infty))log(\frac{b}{a})

    Would love to know where you got this question from
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    (Original post by DFranklin)
    Replacing \infty by N (in other words, using your argument, but with a finite upper limit that we will later let go to infinity) , you have \displaystyle \int_a^b \left[\dfrac{f(xy)}{y}\right]_0^N \, dy = \displaystyle \int_a^b [f(Ny)-f(0)]\frac{1}{y} \, dy

    But \displaystyle \int_a^b f(Ny)/y\, dy =  \int_{Na}^{Nb} f(x) \frac{1}{x}\,dx (*) (by the linear sub x = Ny).

    The given condition on \int f(x)\frac{1}{x}\,dx existing is enough to show that (*) goes to 0 as N goes to infinity.
    I see, thank you.

    (Original post by Lord Nutter)
    Would love to know where you got this question from
    This was from an example sheet (Vector Calculus).
 
 
 
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