Solving an integral equation using the Fourier transform I have to solve the equation 
$\int_0^{\infty} f(x) \cos{(\alpha x)}\, dx=\frac{\sin{\alpha }}{\alpha}$
Using fourier transform. I know this is half of the usual fourier cosine transform, and so that I would get back $f(x)$ using $\frac{2}{\pi} \int_0^{\infty} \frac{\sin{\alpha} \cos{\alpha x}}{\alpha}d{\alpha}$
Is this correct? How do I do this integral?
 A: $$ \begin{align} \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin (\alpha) \cos (x \alpha)}{\alpha} \ d \alpha &= \frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \big((1+x) \alpha \big)+\sin \big( (1-x) \alpha \big)}{\alpha} \ d \alpha \\ &= \frac{1}{\pi} \Big(\text{sgn}(1+x) \frac{\pi}{2}+\text{sgn}(1-x) \frac{\pi}{2} \Big) \\ &= \frac{1}{2} \Big(\text{sgn}(1+x) + \text{sgn}(1-x) \Big) \\ &= \begin{cases} \frac{1}{2}(-1+1) = 0 & \text{if} \ x <-1 \\ \frac{1}{2}(0+1) = \frac{1}{2} & \text{if} \ x = -1 \\ \frac{1}{2} (1+1) = 1 & \text{if} -1 < x <1 \\ \frac{1}{2}(1+0) = \frac{1}{2} & \text{if} \ x = 1 \\ \frac{1}{2} (1-1) = 0 & \text{if} \ x >1\end{cases} \end{align}$$
A: The RHS of your equation is real and even in $\alpha$.  Therefore, its FT is real and even in $x$, its transform variable.  Therefore, the equation is equivalent to
$$\int_{-\infty}^{\infty} dx \: f(x) e^{i \alpha x} = 2 \frac{\sin{\alpha}}{\alpha}$$
Inverting this transform, we get
$$f(x) = \begin{cases} \\ 1 & |x| < 1\\0 & |x| > 1\end{cases}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{{2 \over \pi}\int_{0}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha} =
{1 \over \pi}\int_{-\infty}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha
\\[5mm] = &\
{1 \over \pi}\,\Re\int_{-\infty}^{\infty}\expo{-\ic \alpha x}\,\,{\sin\pars{\alpha}  \over \alpha}\,\dd\alpha \\[5mm] = &\
{1 \over \pi}\,\Re\int_{-\infty}^{\infty}\expo{-\ic \alpha x}
\pars{{1 \over 2}\int_{-1}^{1}\expo{\ic k\alpha}\dd k}\dd\alpha
\\[5mm] = &\
\Re\int_{-1}^{1}\ \underbrace{
\int_{-\infty}^{\infty}\expo{\ic\pars{k - x}\alpha}\,\,\,
{\dd\alpha \over 2\pi}}_{\ds{\delta\pars{k - x}}}\,\dd k =
\bracks{\verts{x} < 1\vphantom{\Large A}}
\end{align}
The case $\ds{x = \pm 1}$ yields $\ds{\color{red}{1/2}}$ which can be directly derived from the proposed  integral. Then,
\begin{align}
&\bbox[5px,#ffd]{{2 \over \pi}\int_{0}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha} =
\left\{\begin{array}{lcl}
\ds{1} & \mbox{if} & \ds{-1 < x < 1}
\\[1mm]
\ds{1 \over 2} & \mbox{if} & \ds{x = \pm 1}
\\[1mm]
\ds{0} &           & \mbox{otherwise}
\end{array}\right.
\end{align}

