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## Homework Statement

For α > 0, determine u(x) by the inverse Fourier transform

[tex]u(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\ \frac{e^{ikx}}{ik+\alpha}\ dk [/tex]

## Homework Equations

## The Attempt at a Solution

This seemed like a relatively simple residue problem. You just note that the denominator has a pole at αi, rewrite the integrand by multiplying through by -i (in order to get the denominator into the form k - αi), and use the residue theorem. As mentioned, the only pole is a simple one, so

[tex]\mbox{Res}_{k=ai}\ \frac{-ie^{ikx}}{k-αi} = -ie^{i(iα)x} = -ie^{-αx}[/tex]

Since the inverse Fourier transform has that factor of 1/2∏, but the residue theorem requires multiplication by 2∏i, I get

[tex]u(x) = e^{-αx}[/tex]

But that seems wrong. Mathematica gives me

[tex]\sqrt{2\pi}e^{\alpha x}\ H(x)[/tex]

Now I remember from days gone by that the square root prefactor is more of a conventional thing, so I'm not so concerned about that. I'm mildly concerned about the sign of α and much, much more concerned about the appearance of the Heaviside function. Where did that come from?

Thanks!