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Math Help - Another integral problem

  1. #1
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    Another integral problem

    I have:

    I(a)= the integral from 0 to 1 of (x^(a)-x^(b))/(ln(x)) where a>b>-1.

    How would I show that I'(a)=1/(a+1)?
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  2. #2
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    Quote Originally Posted by James234 View Post
    I have:

    I(a)= the integral from 0 to 1 of (x^(a)-x^(b))/(ln(x)) where a>b>-1.

    How would I show that I'(a)=1/(a+1)?
    We have:

    <br />
I(a)= \int_0^1 \frac{x^a-x^b}{\ln(x)}dx= \int_0^1 \frac{x^a}{\ln(x)}dx- \int_0^1 \frac{x^b}{\ln(x)}dx<br />

    so:

    <br />
\frac{d}{da}I(a)= \frac{d}{da}\int_0^1 \frac{x^a}{\ln(x)}dx= \frac{d}{da}\int_0^1 \frac{e^{a \ln (x)}}{\ln(x)}dx<br />

    Now differentiating under the integral gives:

    <br />
\frac{d}{da}I(a)= \int_0^1 \ln(x) \frac{e^{a \ln (x)}}{\ln(x)}dx=\int_0^1 x^a dx = \left[ \frac{x^{a+1}}{a+1} \right]_0^1=\frac{1}{a+1}<br />

    RonL
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