I have the following problem:
The task is to show that $$f^*(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(k) e^{ik(x-vt)} dk$$ with $f(k) = A\cdot e^{-a|k - k_0|}$ equals $$f^*(k) = \frac{A}{\sqrt{2\pi}} \frac{2a}{(x-vt)^2 + a^2}e^{ik_0(x-vt)} $$
What I did was say $\gamma(k) := k - k_0 \Leftrightarrow k = \gamma + k_0$; $\Rightarrow d\gamma = dk$; $\gamma(\pm\infty) = \pm \infty$:
$$f(k)^* = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty A\cdot e^{-a|\gamma|} e^{i(\gamma +k_0)(x-vt)} d\gamma = \frac{A}{\sqrt{2\pi}} e^{ik_0(x-vt)} \int_{-\infty}^\infty e^{-a|\gamma|} e^{i\gamma(x-vt)} d\gamma$$
And that is where I am stuck at the moment. By comparing coefficients I should have to show that:
$$ \int_{-\infty}^\infty e^{-a|\gamma|} e^{i\gamma(x-vt)} d\gamma = \frac{2a}{(x-vt)^2 + a^2}$$
But how should that be possible? I've tried applying the dirac-delta-function somewhere but that did not relly work out... Have I overseen something?
Thank you very much for your help.
FunkyPeanut
You can use the identity that $$\int e^{kx}dx=\frac{1}{k}e^{kx}+C$$ for complex constants $k$ and $C$.
Thus the last integral can be evaluated as follows:- $$\int_{-\infty}^\infty e^{-a|\gamma|} e^{i\gamma(x-vt)} d\gamma = \int_{-\infty}^0 e^{a\gamma} e^{i\gamma(x-vt)} d\gamma+\int_{0}^\infty e^{-a\gamma} e^{i\gamma(x-vt)} d\gamma\\=\int_{-\infty}^0 e^{\gamma [i(x-vt)+a]} d\gamma+\int_{0}^\infty e^{\gamma [i(x-vt)-a]} d\gamma\\=\frac{1}{i(x-vt)+a}[e^{\gamma [i(x-vt)+a]}]_{-\infty}^0+\frac{1}{i(x-vt)-a}[e^{\gamma [i(x-vt)-a]}]_{0}^\infty\\=\frac{1}{i(x-vt)+a}-\frac{1}{i(x-vt)-a}\\=\frac{-2a}{-[a-i(x-vt)][a+i(x-vt)]}\\=\frac{2a}{(x-vt)^2+a^2}$$