Toronto Math Forum
APM3462016F => APM346Tests => TT2 => Topic started by: Victor Ivrii on November 17, 2016, 03:28:26 AM

Find Fourier transforms of the functions
\begin{equation}
f_\pm (x)= e^{\varepsilon x}\theta(\pm x)
\end{equation}
and write these function as a Fourier integrals, where $\theta$ is a Heaviside function: $\theta(t)=1$ for $t>0$ and $\theta(t)=0$ for $t<0$.
Bonus (1pt).
Write Fourier transforms of the functions $g(x)=f_+(x)+ f_(x)$ and $h(x)= f_+(x) f_(x)$.

First of all, I think the description of the problem is a little bit problematic since it does not define $\theta(0)$. I would use the convention that $\theta(0)=0$
Here is a reference: https://en.wikipedia.org/wiki/Heaviside_step_function
For
$${f_ + }(x) = {e^{  \varepsilon x}}\theta ( + x) = \left\{ {\matrix{
{{e^{  \varepsilon x}}} & {x \ge 0} \cr
0 & {x < 0} \cr
} } \right.$$
$${{\hat f}_ + }(k) = {1 \over {2\pi }}\int_0^\infty {{e^{  (\varepsilon + ik)x}}dx = } \left. {  {{{e^{  (\varepsilon + ik)x}}} \over {2\pi (\varepsilon + ik)}}} \right_0^\infty = {1 \over {2\pi (\varepsilon + ik)}}$$
Similarly,
$${{\hat f}_  }(k) = {1 \over {2\pi }}\int_{  \infty }^0 {{e^{(\varepsilon  ik)x}}dx = } \left. {{{{e^{(\varepsilon  ik)x}}} \over {2\pi (\varepsilon  ik)}}} \right_{  \infty }^0 = {1 \over {2\pi (\varepsilon  ik)}}$$
Therefore,
$${f_ + }(x) = \int_{  \infty }^\infty {{1 \over {2\pi (\varepsilon + ik)}}} {e^{ikx}}dk$$
$${f_  }(x) = \int_{  \infty }^\infty {{1 \over {2\pi (\varepsilon  ik)}}} {e^{ikx}}dk$$
And
$$\hat g(x) = {{\hat f}_ + }(k) + {{\hat f}_  }(k) = {1 \over {2\pi (\varepsilon + ik)}} + {1 \over {2\pi (\varepsilon  ik)}} = {{2\varepsilon } \over {2\pi ({\varepsilon ^2} + {k^2})}}$$
$$\hat h(x) = {{\hat f}_ + }(k)  {{\hat f}_  }(k) = {1 \over {2\pi (\varepsilon + ik)}}  {1 \over {2\pi (\varepsilon  ik)}} = {{  2ik} \over {2\pi ({\varepsilon ^2} + {k^2})}}$$

Just a remark: if a function is different or undefined at a single point (or any finite number of points) it does not change the value of the integral. This is because a finite number of points has Jordan measure zero (https://en.wikipedia.org/wiki/Jordan_measure). So there is no problem in not defining $\theta(0)$.

So there is no problem in not defining $\theta(0)$.
Well...I think the main problem is that Fourier transformation is defined on the whole real line. So, if $\theta$ is not defined at $0$, then it may be questionable to do all the later calculation.
Also...the domain of Heaviside function does contain $0$...

If we are talking about functions we do not care about their values in a few particular points (distribution will be another matter but for them "value at some particular point" is not defined). Heaviside function could be defined at $0$ as 0 (to make it semicontinuous from the left), $1$ (to make it semicontinuous from the right), or $\frac{1}{2}$ (as a halfsum of the limits).