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Thread: real value function (integral)

  1. #1
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    real value function (integral)



    How would I do this? Im really stuck. Most appreciated!
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  2. #2
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    $\displaystyle \frac{df}{d \alpha} \: = \: \frac{1}{\alpha}$
    Integrating
    $\displaystyle f(\alpha) \: = \: ln | \alpha | +C$
    $\displaystyle f(1) \: = \: C=0$

    $\displaystyle f(\alpha \beta) \: = \: ln | \alpha | +ln | \beta |$
    $\displaystyle f(\alpha \beta) \: = \: f(\alpha)+f(\beta) $
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  3. #3
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    If you wanted to evaluate it, you could write it as $\displaystyle \int^{\infty}_{0} \int^{\alpha}_{1} e^{-tx} \ dt \ dx $

    then switch the order of integration

    $\displaystyle = \int^{\alpha}_{1} \int_{0}^{\infty} e^{-tx} \ dx \ dt $

    $\displaystyle = \int^{\alpha}_{1} \frac{1}{t} \ dx = \ln \alpha $
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  4. #4
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    thanks! If possible, could you help me in this thread? Thanks

    http://www.mathhelpforum.com/math-he...tive-help.html
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  5. #5
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    No , the question says without evaluating the integral , we cannot do it in this way .


    $\displaystyle f( a) = \int_0^{\infty} \frac{ e^{-x} - e^{-ax} }{x}~dx $

    Sub. $\displaystyle x = bt $ and change the dummy variable , we have

    $\displaystyle f(a) = \int_0^{\infty} \frac{ e^{-bx} - e^{-abx} }{bx}~(bdx)$


    $\displaystyle = \int_0^{\infty} \frac{ e^{-bx} - e^{-abx} }{x}~dx $

    $\displaystyle = \int_0^{\infty} \frac{ e^{-bx} - e^{-x} + e^{-x} - e^{-abx} }{x}~dx $


    $\displaystyle =\int_0^{\infty} \frac{ (e^{-x} - e^{-abx})-(e^{-x} - e^{-bx} ) }{x}~dx $

    $\displaystyle = f(ab) - f(b) $ so $\displaystyle f(ab) = f(a) + f(b)$
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