Prove Complex Integral: $\int_m^\infty\sqrt{x^2-m^2}e^{-ixt}dx$

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SUMMARY

The integral $\int_m^\infty\sqrt{x^2-m^2}e^{-ixt}dx$ asymptotically approaches $e^{-imt}$ as $t$ approaches infinity. The proof involves substituting $x = y + m$, which transforms the integral into a form that includes the Inverse Fourier Transform of the function $f(x) = \sqrt{y(y+2m)}$. The key step is to demonstrate that the limit $\lim_{t \to \infty} F(t) = 1$, where $F(t)$ represents the transformed function.

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  • Understanding of complex integrals and asymptotic analysis.
  • Familiarity with Fourier Transforms and their inverses.
  • Knowledge of substitution techniques in calculus.
  • Basic concepts of limits in mathematical analysis.
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  • Explore asymptotic analysis techniques for complex integrals.
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Silviu
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Hello! I found a proof in my physics books and at a step it says that: ##\int_m^\infty\sqrt{x^2-m^2}e^{-ixt}dx \sim_{t \to \infty} e^{-imt}##. Any advice on how to prove this?
 
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Hi @Silviu:

Here is a thought that might help.

Substitute x= y+m. This will produce e-imt F(t)
where
F(t) = Inverse Fourier Transform of f(x)
where
f(x) = √(y (y+2m)).
Therefore, this reduces the problem to prove lim[t→∞] F(t) = 1.

Good luck.

Regards,
Buzz
 

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