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Math Help - Improper integral

  1. #1
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    Improper integral

    Evaluate

    \int_0^\infty\frac{\cos x-e^{-x}}x\,dx
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  2. #2
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    \frac1x=\int_0^\infty e^{-ux}\,du.

    You can do the rest from there, it's pretty straightforward.

    The answer is 0.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Okay, let's finish this.

    Quote Originally Posted by liyi View Post
    Evaluate

    \int_0^\infty\frac{\cos x-e^{-x}}x\,dx
    \begin{aligned}<br />
\int_0^\infty\frac{\cos x-e^{-x}}x\,dx&=\int_0^\infty\int_0^\infty e^{-ux}\left(\cos x-e^{-x}\right)\,du\,dx\\<br />
&=\int_0^\infty\int_0^\infty\Big[e^{-ux}\cos x-e^{-(u+1)x}\Big]\,dx\,du\\<br />
&=\int_0^\infty\left(\frac u{u^2+1}-\frac1{u+1}\right)\,du\\<br />
&=\left.\ln\left|\frac{\sqrt{u^2+1}}{u+1}\right|\r  ight|_0^\infty\\<br />
\end{aligned}

    The conclusion follows.
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