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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.

Where is my brain-overflow in the calculation given?

Thank you in advance


PS.: Even though it is not such a thing, I marked this question as homework since it is on this level.

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up vote 3 down vote accepted

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.)

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Thank you for this answer and the additional information. I think there was a brain overflow in understanding where the $\delta$ comes from :) Greets – Robert Filter Jan 19 '11 at 16:47

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 $\gamma_2 e^{\gamma_2 z}$

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@loenbloy: Thank you for your fast response. I think the three given ones make a pretty decent picture of the situation, all from a slightly different perspective :) Greets – Robert Filter Jan 19 '11 at 16:49

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 $e^{-ik0}=1$, $$ \int_{ - \infty }^\infty {e^{ - ikz} dw(z)} = \int_{ - \infty }^0 {e^{ - ikz} w(z)dz} + (-\gamma_1 - \gamma_2) + \int_{ 0 }^\infty {e^{ - ikz} w(z)dz}. $$

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Thank you for your answer. I must admit that I never had any course on measure theory so I have to speculate that I really want to calculate the first version. Greets – Robert Filter Jan 19 '11 at 16:20
The first version corresponds to your solution; the second version corresponds to a contribution of $-\gamma_1 - \gamma_2$. – Shai Covo Jan 19 '11 at 16:30
Thank you again for the clarification! I now know that there is a difference in these two approaches :) Greets – Robert Filter Jan 19 '11 at 16:45
Thank you for the further clarification. – Robert Filter Jan 20 '11 at 7:55

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