Applying the inversion theorem, we may write
$$h(x) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{i\xi x}\sqrt{\frac{2}{\pi}}\frac{\sin(\xi)}{\xi} \mathrm{d}\xi$$
which leads to
$$\int_{-\infty}^{\infty} \frac{1}{\pi}\frac{\sin(\xi)}{\xi} \mathrm{d}\xi = h(0) = 1. $$
This in turn implies
$$\bbox[5px,border:2px solid #CAAA00]{\int_{0}^{\infty} \frac{\sin(x)}{x} \mathrm{d}x = \frac{\pi}{2}}$$
To calculate the second integral it is sufficient to use Plancherel. You should get $\pi/2$ as well if I remember correctly. The third one can be calculated in a similar fashion:
$$\int_{-\infty}^{\infty} \left(\sqrt{\frac{2}{\pi}}\frac{\sin(\xi)} {\xi}\right)^4 \mathrm{d}\xi =\|\hat{h}^2\|_{L^2}^2 = \frac{1}{2\pi} \|\widehat{h\ast h}\|_{L^2}^2 = \frac{1}{2\pi}\|h \ast h\|_{L^2}^2 = \frac{1}{2\pi}\frac{16}{3} $$
so we obtain
$$\bbox[5px,border:2px solid #CAAA00]{\int_{0}^{\infty}\frac{\sin^4(x)}{x^4} \mathrm{d}x = \frac{\pi}{3}}$$
The solution is $$f(y)=\frac{2}{\pi}\frac{y}{1+y^2}$$ I will add the derivation soon.
Derivation:
Let
$$g(y)=
\begin{cases}
f(y)&&y\ge0 \\
0&&y<0
\end{cases}
$$
Then, with Euler's formula, we can rephrase the problem as
$$\int^\infty_{-\infty}g(y)\left(\frac{e^{ixy}-e^{-ixy}}{2i}\right)dy=e^{-x}$$
Equivalently,
$$\mathcal F\{g(y)\}(-x)-\mathcal F\{g(y)\}(x)=i\sqrt{\frac2\pi}e^{-x}$$
$$G(-x)=G(x)+i\sqrt{\frac2\pi}e^{-x}\qquad{x>0}$$
This functional equation does not tell much, as the solution is not unique. The best we can do is defining
$$G(x)=
\begin{cases}
\varphi (x) && x>0 \\
\varphi(-x)+i\sqrt{\frac2\pi}e^{x} && x<0
\end{cases}
$$
for some $\varphi (x):\mathbb R^+$ with sufficiently nice properties.
Then,
$$
\begin{align}
g(y)&=\frac{1}{\sqrt{2\pi}}\int^\infty_{-\infty}G(x)e^{ixy}dx \\
\sqrt{2\pi}g(y)&=\int^\infty_{0}G(x)e^{ixy}dx+\int^{\infty}_{0}G(-x)e^{-ixy}dx \\
&=\int^\infty_{0}\varphi(x)e^{ixy}dx+\int^{\infty}_{0}\left(\varphi(x)+i\sqrt{\frac2\pi}e^{-x}\right)e^{-ixy}dx \\
&=2\int^\infty_{0}\varphi(x)\cos(xy)dx+i\sqrt{\frac2\pi}\int^{\infty}_{0}e^{-x}e^{-ixy}dx \\
&=2\int^\infty_{0}\varphi(x)\cos(xy)dx+i\sqrt{\frac2\pi}\frac1{1+iy} \\
g(x)&=\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(xy)dx+\frac i\pi\frac1{1+iy} \\
\end{align}
$$
Let $a>0$. By assumption, $g(-a)=0$. Therefore,
$$\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(ay)dx+\frac i\pi\frac1{1-ia}=0$$
Then,
$$\begin{align}
g(a)&=\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(ay)dx+\frac i\pi\frac1{1+ia} \\
&=-\frac i\pi\frac1{1-ia}+\frac i\pi\frac1{1+ia} \\
&=\frac2\pi\frac{a}{1+a^2}
\end{align}
$$
Hence, $$f(y)=\frac{2}{\pi}\frac{y}{1+y^2}$$
Verification:
(The calculations below are a bit sloppy as I assumed that the integral and differentiation can be interchanged.)
$$\begin{align}
L.H.S.
&=\frac2\pi \int^\infty_0\frac{y}{1+y^2}\sin(xy)dy \\
&=-\frac2\pi\frac{\partial}{\partial x}\int^\infty_0\frac{\cos(xy)}{1+y^2}dy \\
&=-\frac1\pi\frac{\partial}{\partial x}\int^\infty_{-\infty}\frac{\cos(xy)}{1+y^2}dy \\
&=-\frac1\pi\frac{\partial}{\partial x}\pi e^{-x} \\
&=e^{-x}\\
&=R.H.S.
\end{align}
$$
Here we utilized the well-known integral identity
$$\int^\infty_{-\infty}\frac{\cos(ax)}{1+x^2}dx=\pi e^{-|a|}$$
For its proof, see the accepted answer here.
Best Answer
This will be clear to those who have dealt with Laplace transforms. The Laplace transform of $\sin wt$ is given by ${w \over w^2 + s^2}$. In other words $$\int_0^{\infty} e^{-st} \sin wt \,dt = {w \over w^2 + s^2}$$ Thus in your notation, the answer to your question is ${a \over w} e^{-ax}$.