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Laplace Transform Integral

  1. Dec 31, 2013 #1
    1. The problem statement, all variables and given/known data

    $$ \int_0^\infty \frac{\sin xt}{x} \, dt $$

    2. Relevant equations

    3. The attempt at a solution

    $$ = \int_0^\infty L(\sin xt) \, dp $$

    $$ = \int_0^\infty \frac{x}{p^2 + x^2} \, dp $$

    $$ = x \int_0^\infty \frac{dx}{p^2 + x^2} \, dp $$

    p = x tan theta:

    $$ = x \int_0^{\pi/2} \frac{ \sec^2 \theta}{x^2 \sec^2 \theta} \, d\theta $$

    $$ = \frac{1}{x} \cdot \frac{\pi}{2} $$

    My textbook says that the answer should be exactly pi /2. What did I do wrong?
    Last edited: Dec 31, 2013
  2. jcsd
  3. Dec 31, 2013 #2
    Never mind, I found the problem: I forgot to include the x in dx = x sec^2 theta d theta. However, while we're here, I have another textbook problem:

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

    I have noticed that this is expressible as

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

    Is that the right first step? I'm not sure where to go from here
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