fourier transform pair From Wikipedia, I get the following fourier transform pairs:
$$\frac{d}{dx}f(x) \rightarrow i2\pi\xi \hat{f}(\xi)$$ and
$$\frac{sgn(x)}{2} \rightarrow \frac{1}{i2\pi\xi}$$
When the right hand side of these equations are multiplied, I get $\hat{f}(\xi)$.
Translating this to left hand side, I expect the convolution of $\frac{d}{dx}f(x)$ and $\frac{sgn(x)}{2}$ to be equal to $f(x)$ (because the RHS is $\hat{f}(\xi)$).
However, if I expand $$\frac{d}{dx}f(x) * \frac{sgn(x)}{2} = \frac{1}{2}\bigg(\int\limits_{-\infty}^{x}\frac{d}{dy}(f(y)) dy - \int\limits_{x}^{\infty}\frac{d}{dy}(f(y)) dy)\Bigg)$$
this equals $f(x) + \frac{f(-\infty)\ +\ f(\infty)}{2}$
Questions 


*

*Are the starting fourier transform pairs correct or is there some
missing factor? OR

*Is there an error in the steps that I have used for convolution? OR

*Is there some property due to which the extra factor of $\frac{f(-\infty)\ +\ f(\infty)}{2}$ become equal to 0? 

 A: There is no mistake in your reasoning. You are just forgetting that, in order for your statements to make sense, you must specify the space of functions on which you choose to define the Fourier transform.
An often used choice is to define it on the space of rapidly decreasing functions (the "Schwartz functions"), thus turning the Fourier transform into an automorphism of this space. These have the property, among others, of vanishing at infinity, so for them $f(-\infty) = f(\infty) = 0$. The problem is that $\operatorname{sgn}$ is not such a function, tending to $1$ at $\infty$.
To cure this, another choice is to use tempered distributions. I guess that this is what you want. In this context your equality will be true.
There are several other choices (most of them including the space $L^1$), but tempered distributions are the most convenient setting.
A: The transforms you have given above assume that the function $f$ decays to $0$ at $\pm \infty$, which would send your troubled term to $0$. Indeed, to show that 
$$
\int_{-\infty}^{\infty} f'(x) e^{-2\pi i x \xi} \,dx = 2 \pi i \xi \hat{f}(\xi)
$$ 
you perform an integration-by-parts on the left-hand integral, moving the derivative from $f$ to the Fourier kernel. However, when you do this (or any) integration-by-parts, you need to worry about the boundary terms, which are zero in the right-hand term.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\,\mathcal{#1}}
 \newcommand{\mrm}[1]{\,\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Notation:
  
   1. $\ds{\mrm{f}\pars{x} =
\int_{-\infty}^{\infty}\hat{\mrm{f}}\pars{k}\expo{-\ic kx}
\,{\dd k \over 2\pi}\,,\qquad\hat{\mrm{f}}\pars{x} =
\int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{\ic kx}\,\dd k}$.
  
  2. $\ds{\Theta: \mbox{Heaviside Step Function}}$.
  $\ds{\mrm{sgn}\pars{x} = \Theta\pars{x} - \Theta\pars{-x}}$.

\begin{align}
\Theta\pars{x} & = -\int_{-\infty}^{\infty}{\expo{-\ic kx} \over k + \ic 0^{+}}
\,{\dd k \over 2\pi\ic}
\quad\imp\quad\hat{\Theta}\pars{k} = {\ic \over k + \ic 0^{+}}
\end{align}

\begin{align}
\mrm{sgn}\pars{x} & =
-\int_{-\infty}^{\infty}{\expo{-\ic kx} \over k + \ic 0^{+}} +
\int_{-\infty}^{\infty}{\expo{\ic kx} \over k + \ic 0^{+}}
\,{\dd k \over 2\pi\ic} =
-\int_{-\infty}^{\infty}{\expo{-\ic kx} \over k + \ic 0^{+}} -
\int_{\infty}^{-\infty}{\expo{-\ic kx} \over -k + \ic 0^{+}}
\,{\dd k \over 2\pi\ic}
\\[5mm] & =
\int_{-\infty}^{\infty}\pars{-\,{1 \over k + \ic 0^{+}} -
{1 \over k - \ic 0^{+}}}\expo{-\ic kx}
\,{\dd k \over 2\pi\ic}
\\[5mm] 
\imp\quad & \hat{\mrm{sgn}}\pars{k} =
{\ic \over k + \ic 0^{+}} + {\ic \over k - \ic 0^{+}}
\end{align}


$\ds{\pm\ic 0^{+}}$ are understood 'under the integral sign'. For instance:
  $$
\int_{-\infty}^{\infty}{\phi\pars{k} \over k + \ic 0^{+}}\,\dd k
\quad\mbox{means}\quad 
\lim_{\epsilon \to 0^{+}}
\int_{-\infty}^{\infty}{\phi\pars{k} \over k + \ic\epsilon}\,\dd k
$$
  and it serves to the purpose of a shortcut in an 'operational side'.


By the way;
$$
\totald{}{x}\bracks{\mrm{f}\pars{x}\,{\mrm{sgn}\pars{x} \over 2}} =
\half\,\mrm{f}'\pars{x}\mrm{sgn}\pars{x} + \mrm{f}\pars{0}\delta\pars{x}
$$
