Laplace Inversion: Why Contour Must Exceed Singularities

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SUMMARY

The discussion centers on the necessity of selecting a contour for the Laplace inversion formula such that the real part, denoted as σ, exceeds all singularities of the function F(s). This requirement ensures that the integral converges and that the resulting function f(t) vanishes for t < 0. The mathematical expression for the Laplace inversion is given as (\mathcal{L}^{-1} F)(t) = \frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} e^{st} F(s) ds, highlighting the importance of proper contour selection in complex analysis.

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bruno67
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Maybe I am just being stupid, but I don't understand why in the Laplace inversion formula

(\mathcal{L}^{-1} F)(t) = \frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} e^{st} F(s) ds
the contour of integration must be chosen so that \sigma is greater than the real part of all singularities of F(s). I would be very grateful if someone could explain this.
 
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Is it so that f(t) vanishes for t<0?
 

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