Proving inverse Fourier transform of 1/(1+x^2) = 1/(1+x^2)

In summary, the conversation discusses solving the integral 1/sqrt(2π) ∫F(t)eitxdt for F(t) = sqrt(π/2)e-t for t>0 and F(t) = sqrt(π/2)et for t<0, and showing that it equals 1/(1+x2). The suggested approach is to break up the real line into two parts and do separate integrals for each part.
  • #1
Vitani11
275
3

Homework Statement


F(t) = sqrt(π/2)e-t for t>0
F(t) = sqrt(π/2)et for t<0
In other words the question asks to solve this integral: 1/sqrt(2π) ∫F(t)eitxdt and show that it equals 1/(1+x2)

Homework Equations


F(t) = sqrt(π/2)e-t for t>0
F(t) = sqrt(π/2)et for t<0
1/sqrt(2π) ∫F(t)eitxdt

The Attempt at a Solution


I want to use complex numbers but these functions are analytic and so I can not do this using residues. A naive approach to integrating with respect to t is straightforward but the limits cause the result to be infinity. How should I approach this properly?
 
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  • #2
Vitani11 said:

Homework Statement


F(t) = sqrt(π/2)e-t for t>0
F(t) = sqrt(π/2)et for t<0
In other words the question asks to solve this integral: 1/sqrt(2π) ∫F(t)eitxdt and show that it equals 1/(1+x2)

Homework Equations


F(t) = sqrt(π/2)e-t for t>0
F(t) = sqrt(π/2)et for t<0
1/sqrt(2π) ∫F(t)eitxdt

The Attempt at a Solution


I want to use complex numbers but these functions are analytic and so I can not do this using residues. A naive approach to integrating with respect to t is straightforward but the limits cause the result to be infinity. How should I approach this properly?
Break up the real line into ##-(\infty, 0)## and ##(0,\infty)##, then do two separate integrals. Each integral is really easy.
 

Related to Proving inverse Fourier transform of 1/(1+x^2) = 1/(1+x^2)

1. How do you prove the inverse Fourier transform of 1/(1+x^2) is equal to 1/(1+x^2)?

The inverse Fourier transform of a function f(x) is given by F^(-1){f(x)} = (1/2π) ∫F(k)e^(ikx) dk, where F(k) is the Fourier transform of f(x). In this case, the Fourier transform of 1/(1+x^2) is given by F(k) = √(π/2)e^(-|k|), which can be derived using the properties of the Fourier transform. Substituting this into the inverse Fourier transform formula, we get F^(-1){F(k)} = (1/2π) ∫√(π/2)e^(-|k|)e^(ikx) dk = 1/(1+x^2). Therefore, the inverse Fourier transform of 1/(1+x^2) is equal to 1/(1+x^2).

2. How does the Fourier transform of 1/(1+x^2) relate to its inverse Fourier transform?

The Fourier transform and inverse Fourier transform are closely related and can be thought of as two sides of the same coin. The Fourier transform converts a function from the time domain to the frequency domain, while the inverse Fourier transform converts it back from the frequency domain to the time domain. In this case, the Fourier transform of 1/(1+x^2) is given by F(k) = √(π/2)e^(-|k|), and its inverse Fourier transform is F^(-1){F(k)} = 1/(1+x^2). Thus, the Fourier transform and inverse Fourier transform are inverse operations of each other in this case.

3. Can you prove that the Fourier transform of 1/(1+x^2) is a real-valued function?

Yes, the Fourier transform of 1/(1+x^2) is a real-valued function. This can be seen by calculating the complex conjugate of the Fourier transform F(k) = √(π/2)e^(-|k|). The complex conjugate of a function is given by F*(k) = √(π/2)e^(|k|), which is equal to F(k). This proves that the Fourier transform of 1/(1+x^2) is a real-valued function.

4. Is the inverse Fourier transform of 1/(1+x^2) unique?

Yes, the inverse Fourier transform of 1/(1+x^2) is unique. This is because the Fourier transform and inverse Fourier transform are one-to-one operations, meaning that each function has a unique Fourier transform and inverse Fourier transform. In this case, the inverse Fourier transform of 1/(1+x^2) is 1/(1+x^2) and there is no other function that can be its inverse Fourier transform.

5. How can the inverse Fourier transform of 1/(1+x^2) be used in practical applications?

The inverse Fourier transform of 1/(1+x^2) has many practical applications, particularly in signal processing and image reconstruction. In signal processing, it can be used to filter out noise from a signal by removing high frequency components. In image reconstruction, it can be used to reconstruct an image from its Fourier transform, which can be useful in medical imaging and astronomy. It is also used in solving differential equations and in quantum mechanics.

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