About Fourier transforms of piecewise linear functions. Consider a function $f$ which is $0$ for $x< 1$ and is say $x-1$ for $x >1$.
Consider a function $g$ which is $0$ for $x <2$ and is say $x -2$ for $x>2$. 
Now using some kind of regulator one can define the Fourier transform for $f$ and $g$ and for both of them the $t^{th}$ frequency will get a term proportional to $\frac{1}{t^2}$.  
From this point of view it seems that the Fourier transform of $f+g$ has its $t^{th}$ mode proportional to $\frac{1}{t^2}$.
But in reality the function $f+g$ has a line segment of equation $x-1$ for $x \in [1,2]$ and this contributes a $\frac{1}{t}$ term to the Fourier transform of the sum. The sum is $0$ for $x<1$ and is $2x-3$ for $x>2$ and these pieces will contribute $0$ and $\frac{1}{t^2}$ (respectively) proportional term in the $t^{th}$ frequency. So this $\frac{1}{t}$ term doesn't get cancelled by any other parts. 
Can someone kindly help resolve this confusion? 
What is the correct Fourier tranform of $f$ , $g$ and $f+g$? 
 A: Fourier transform definition:
$$F(t)=\int_{-\infty}^{\infty}e^{ixt}f(x)dx$$
the choice $f(x)=(x-b)\theta(x-a)$ gives
$$F(t)=i(a-b)e^{iat}\frac{1}{t}-e^{iat}\frac{1}{t^2}-\pi b \delta (t)-i \pi \delta '(t)\qquad(*)$$
so the Fourier transform of $(2x-3)\theta(x-2)$, corresponding to $a=2$, $b=3/2$, does have a term proportional to $1/t$, unlike what was written in the OP. That should resolve this confusion, I hope.

I see the confusion remains. Let me add a bit more, in support of the expression given above. It seems we agree on the Fourier transforms of $f_1(x)=\theta(x)$ and of $f_2(x)=x\theta(x)$, with the proper distributional interpretation:
$$F_1(t)=it^{-1}+\pi\delta(t),\;\;F_2(t)=-t^{-2}-i\pi\delta'(t)$$
Direct integration gives you the Fourier transform of $f_3(x)=(x-b)\theta(a-x)\theta(x)$, which is nonzero only in the interval $0\lt x\lt a$:
$$F_3(t)=\frac{e^{i a t} (1-i a t+i b t)-i b t-1}{t^2}$$
The desired $F(t)$ is then the Fourier transform of $f(x)=f_2(x)-bf_1(x)-f_3(x)$, and linearity of the Fourier transform gives you the formula $(*)$ above, including the terms decaying as $1/t$.

As a curiosity, I might add that Mathematica is confused about these issues as well ... 
