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Math Help - Prove directly

  1. #1
    Newbie
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    Prove directly

    How to prove the following equality directly without assuming it represents the natural log function?
    \displaystyle \int_{1}^{yz} t^{-1}\, dt=\int_{1}^{y} t^{-1}\, dt+\int_{1}^{z} t^{-1}\, dt.
    Thanks!
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  2. #2
    Junior Member bondesan's Avatar
    Joined
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    Proposition:

    \displaystyle{\int_1^x \frac{1}{t}dt = \int_a^{ax}\frac{1}{t}dt}

    Proof: Let u=at\Rightarrow dt=\frac{du}{a}. If t=1, then u=a and if t=x, u=ax. Therefore:

    \displaystyle{\int_1^x\frac{1}{t}dt=\int_a^{ax}\fr  ac{1}{u/a}\cdot\frac{du}{a} = \int_a^{ax}\frac{1}{u}du }

    This show us that this integral is invariant by translation.

    So \displaystyle{\int_1^{yz} \frac{1}{t}dt = \int_1^{y}\frac{1}{t}dt + \int_y^{yz} \frac{1}{t}dt}.

    By the proposition you get the answer.
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