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Math Help - Integral

  1. #1
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    Integral

    Evaluate \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

    \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:

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

    \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

    \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

    \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:

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

    \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

    \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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