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Math Help - Today's calculation of integral #4

  1. #1
    Math Engineering Student
    Krizalid's Avatar
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    Today's calculation of integral #4

    No matter what k is. Let <br />
\alpha=\int_0^1\frac{e^u}{1+u}\,du<br />
& <br />
\beta=\int_{k-1}^k\frac{e^{-x}}{x-k-1}\,dx.<br />
Assume that <br />
\lambda\cdot \alpha=\beta.<br />
Compute \lambda.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Krizalid View Post
    No matter what k is. Let <br />
\alpha=\int_0^1\frac{e^u}{1+u}\,du<br />
& <br />
\beta=\int_{k-1}^k\frac{e^{-x}}{x-k-1}\,dx.<br />
Assume that <br />
\lambda\cdot \alpha=\beta.<br />
Compute \lambda.
    let -x=u-k \mbox{ then } dx=-du

    <br />
\beta=\int_{k-1}^k\frac{e^{-x}}{x-k-1}\,dx=\int_{1}^{0}\frac{e^{u-k}}{-u-1}(-du)=\frac{1}{e^k}\int_{0}^{1}\frac{e^u}{u+1}du<br />

    so \lambda =\frac{\beta}{\alpha}=\frac{\frac{1}{e^k}\int_{0}^  {1}\frac{e^u}{u+1}du}{\int_{0}^{1}\frac{e^u}{u+1}d  u}=\frac{1}{e^k}
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    That's correct.
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