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Laplace Transform Integral

  1. Jan 1, 2014 #1
    I posted this in the homework section, but I haven't received any help, so hopefully putting it in this section won't be an issue. I'm trying to compute the integral

    $$ \int_0^ \infty \frac{ \cos xt}{1 + t^2} \, dt $$

    using the Laplace transform. The first thing that catches my eye is the 1 /(1 + t^2) factor, which is equal to the Laplace transform of sin x:

    $$ = \int_0^\infty \cos xt \, L[ \sin x ] \, dt $$

    Any ideas?
     
  2. jcsd
  3. Jan 2, 2014 #2

    vanhees71

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    Why do you want to use the Laplace transformation here? I'd rather use the Fourier transformation. It's simpler to put it first in the exponential form. Your Integral is
    [tex]F(x)=\frac{1}{2}\int_0^{\infty} \frac{\exp(\mathrm{i} x t)+\exp(-\mathrm{i} x t)}{1+t^2}.[/tex]
    Substituting [itex]t'=-t[/itex] in the second integral you get after some algebra
    [tex]F(x)=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\exp(\mathrm{i} x t)}{1+t^2}. \qquad (1)[/tex]
    Now we use the fact that
    [tex]\int_{-\infty}^{\infty} \frac{\mathrm{d} x}{2 \pi} \exp(-|x|) \exp(-\mathrm{i} t x)=\frac{1}{\pi (1+t^2)}.[/tex]
    Thus (1) is (up to a factor [itex]\pi/2[/itex]) the inverse of this Fourier transform. This gives
    [tex]F(x)=\frac{\pi \exp(-|x|)}{2}.[/tex]
     
  4. Jan 2, 2014 #3
    Ok, thanks for that. Do you also see any way to do it with the Laplace transform?
     
  5. Jan 3, 2014 #4

    vanhees71

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    Hm, I've no idea. Perhaps you can somehow use the convolution theorem with some clever trick?
     
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