What is the Laplace Transform Integral for the function $\frac{\cos xt}{1+t^2}$?

[pierce15]

Homework Statement

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

Homework Equations

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?

 

[pierce15]

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