Fourier Transform of an unsteady function I am a little stuck at the moment with the Fourier transform of a function $w(z)$ that is defined piecewise and discontinuous at $z = 0$:
$$w\left(z\right) = \begin{cases} -\gamma_1 e^{-\gamma_{1}z} & z > 0 \\ \gamma_2 e^{\gamma_{2}z} & z < 0\end{cases}$$
with $\gamma_i \geq 0$.
In a naive attempt, I calculated the following:
$$ W \left( k \right) = \int_{-\infty}^{\infty}w(z) e^{-\mathrm{i}k z}dz$$
$$ =  -\gamma_{1}\int_{0}^{\infty}e^{-\left(\gamma_{1}+\mathrm{i}k\right)z}dz+\gamma_{2}\int_{-\infty}^{0}e^{\left(\gamma_{2}-\mathrm{i}k\right)z}dz  = -\frac{\gamma_{1}}{\left(\gamma_{1}+\mathrm{i}k\right)}+\frac{\gamma_{2}}{\left(\gamma_{2}-\mathrm{i}k\right)}$$
But I am almost sure that I miscalculated since there should somewhere occur a $\delta$-contribution proportional to $\left(\gamma_2 - \gamma_1 \right)\cdot\delta(0)$ as in the case of the Heaviside function.
Hence:

Where is my brain-overflow in the calculation given? 

Thank you in advance
Sincerely
Robert
PS.: Even though it is not such a thing, I marked this question as homework since it is on this level. 
 A: There's no brain overflow. The $\delta(x)$ part in the Heaviside calculuation comes from the fact that the Heaviside function does not decay at infinity. As $x\to +\infty$, you have $H(x)$ is bounded from below by a constant.
The fact that your function has an exponentially decreasing weight gets rid of that problem. (When the weight $\gamma_i = 0$, by definition your function is zero also, so it has zero Fourier transform.)
Key thing to remember when looking at Fourier transforms: it interchanges decay properties at infinity with smoothness of a function. That is (morally speaking), if your function is smooth, then its Fourier transform will have fast decay near infinity. If your function has exponential decay, then its Fourier transform will be smooth. 
That your starting function has a jump discontinuity at the origin is thus reflected in the roughly $1/k$ (slow) decay of the Fourier transform (analogous to the second factor in the Fourier transform of the Heaviside function in your link). 
(Also see rule 205 in Wikipedia's list.)
A: It depends on whether you want to compute $\int_{ - \infty }^\infty  {e^{ - ikz} w(z)dz} $ or $\int_{ - \infty }^\infty  {e^{ - ikz} dw(z)} $.
Further clarification: Define $w(0)=-\gamma_1$, so that $w$ is right-continuous with left limits. Since $w$ has a jump discontinuity of size $-\gamma_1 - \gamma_2$ at $z=0$, and since $e^{-ik0}=1$,
$$
\int_{ - \infty }^\infty  {e^{ - ikz} dw(z)} = \int_{ - \infty }^0  {e^{ - ikz} w(z)dz} + (-\gamma_1 - \gamma_2) + \int_{ 0 }^\infty  {e^{ - ikz} w(z)dz}.
$$
A: You should ask yourself what happens with the evaluation of the integral primitive at infinity (when applying the Barrow rule) if $\gamma =0$. For example, what happens to $e^{-(\gamma_1 + ik)z}$ evaluated in $z=\infty$ if $\gamma_1 =0$?
That case must be treated with some care. But, in your case, it's no problem, because then your function is zero and the result is ok. Then, I think your result is correct (but you must go with care, and treat the cases $\gamma_i =0$ separatedly).
It would be different is, for example, your second piece of function would have been $e^{\gamma_2 z}$ instead of $\gamma_2 e^{\gamma_2 z}$
