Question

Solve the integral equation given by $f(x) + 4 \int_{-\infty}^{\infty} e^{-a|x-t|} f(t) dt = g(x)$ (5) using Fourier transforms. What is the closed form for the solution when $a = 1$ and $g(x) = e^{-|x|}$? Hint: use contour integration.

          Solve the integral equation given by
$f(x) + 4 \int_{-\infty}^{\infty} e^{-a|x-t|} f(t) dt = g(x)$ (5)
using Fourier transforms. What is the closed form for the solution when $a = 1$ and $g(x) = e^{-|x|}$? Hint:
use contour integration.

Solve the integral equation given by
f(x) + 4 ∫-∞^∞ e^-a|x-t| f(t) dt = g(x) (5)
using Fourier transforms. What is the closed form for the solution when a = 1 and g(x) = e^-|x|? Hint:
use contour integration.

Added by Brian T.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Aishwarya Krishnakumar

Step 1

The Fourier transform of (x)^6 is given by: F[(x)^6] = ∫[from -∞ to ∞] (x)^6 e^(-2πixω) dx Using the property of the Fourier transform, we have: F[(x)^6] = (i/2π) * d^6/dω^6 [∫[from -∞ to ∞] e^(-2πixω) dx] The integral in the above expression is the Fourier Show more…

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Please solve showing steps. Thank you very much!!! Solve the integral equation given by (x)^6 = 7p(x)f(|2-x|p) (5) using Fourier transforms. What is the closed form for the solution when a = 1 and g(x) = e^(-|x|)? Hint: use contour integration.