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

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SUMMARY

The discussion centers on evaluating the Laplace Transform Integral for the function $\frac{\cos xt}{1+t^2}$. The integral is expressed as $\int_0^\infty \frac{\cos xt}{1+t^2} \, dt$, which can be approached using the Laplace Transform of the sine function. The participant initially miscalculated an integral involving $\sin xt$ but later identified the omission of a factor in their differential substitution. The correct evaluation leads to a solution involving the Laplace Transform, confirming that the integral converges to a specific value.

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  • Understanding of Laplace Transforms
  • Familiarity with trigonometric integrals
  • Knowledge of substitution techniques in calculus
  • Experience with improper integrals
NEXT STEPS
  • Study the properties of the Laplace Transform, particularly for sine and cosine functions.
  • Learn about integration techniques involving trigonometric functions and their transforms.
  • Explore the derivation of the Laplace Transform for functions of the form $\frac{\cos xt}{1+t^2}$.
  • Investigate common pitfalls in integral calculus, especially in substitution methods.
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Students of calculus, mathematicians working with integral transforms, and anyone seeking to deepen their understanding of Laplace Transforms and trigonometric integrals.

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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?
 
Last edited:
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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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