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Thread: Integral

  1. #1
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    Integral

    Evaluate $\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx$
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  2. #2
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    Krizalid's Avatar
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    Fact you should know

    $\displaystyle \int_0^\infty e^{-x}\cos(\alpha x)\,dx=\frac1{\alpha^2+1}$

    Now, let's introduce the following parameter to construct a double integral as follows:

    $\displaystyle \int_0^1\cos(\alpha x)\,d\alpha=\frac{\sin x}x,$ this yields

    $\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^\infty e^{-x}\left(\int_0^1\cos(\alpha x)\,d\alpha\right)\,dx$

    Reversing the order of integration

    $\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^1\int_0^\infty e^{-x}\cos(\alpha x)\,dx\,d\alpha=\int_0^1\frac1{\alpha^2+1}\,d\alph a=\frac\pi4\,\blacksquare$
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Fact you should know

    $\displaystyle \int_0^\infty e^{-x}\cos(\alpha x)\,dx=\frac1{\alpha^2+1}$

    Now, let's introduce the following parameter to construct a double integral as follows:

    $\displaystyle \int_0^1\cos(\alpha x)\,d\alpha=\frac{\sin x}x,$ this yields

    $\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^\infty e^{-x}\left(\int_0^1\cos(\alpha x)\,d\alpha\right)\,dx$

    Reversing the order of integration

    $\displaystyle \int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^1\int_0^\infty e^{-x}\cos(\alpha x)\,dx\,d\alpha=\int_0^1\frac1{\alpha^2+1}\,d\alph a=\frac\pi4\,\blacksquare$
    Impressive, as always!

    -Dan
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