How do I solve the integral equation $\int_0^\infty f(y)\sin(xy)dy=e^{-x} , x>0$? 
I'm trying to solve this integral equation, but I can't deal with the sine!
  $$\int_0^\infty f(y)\sin(xy)dy=e^{-x} , x>0$$

How do I proceed?
I tried using the fourier transform but then I can't deal with the fourier transform of the sine.
The definition of Fourier transform I'm using is:
$$f(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i\xi x}dx$$
EDIT: I think I know what my problem was, I was doing the fourier transform of the sine individualy instead of doing the fourier transform of $f(y)\times sin(xy)$. Doing this and substituing the sine for its exponential formula will give the answer right?
 A: Note that as $x \to 0+$, $\sin(xy) \to 0$ so
if $f$ is integrable, you should have $\int_0^\infty f(y) \sin(xy)\; dy \to 0$ by Dominated Convergence.  But of course $e^{-x}$ goes to $1$, not $0$.  So there is certainly no integrable solution.  
I suppose it's possible that there is a non-integrable solution, but that makes convergence of your integral problematic.  Do you have any reason to expect a solution to exist?
A: 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.
A: If $f_o$ is the odd extension of $f$ to $\mathbb{R}$, then
$$
   e^{-x} = \int_0^{\infty}f(y)\sin(xy)dy \\
   = \frac{1}{2}\int_{-\infty}^{\infty}f_o(y)\sin(xy)dy \\
   = \frac{1}{2i}\int_{-\infty}^{\infty}f_o(y)e^{ixy}dy,
$$
which works out by residues if you close in the upper half plane and you use
$$
         f_o(y)=\frac{1}{\pi}\left(\frac{1}{y-i}+\frac{1}{y+i}\right) = \frac{2}{\pi}\frac{y}{y^2+1}.
$$
As anticipated by Robert Israel, $f(y)$ is not integrable on $(0,\infty)$. The solution is valid when interpreted as an improper integral at $\infty$, which fits nicely with Complex Analysis and with Fourier Analysis. Any real function that is locally integrable on $[0,\infty)$ and is eventually monotonically increasing or decreasing to $0$ has a Fourier transform because the sine and cosine transforms can be interpreted as sums of integrals over a finite interval plus an alternating series with a general term that tends to $0$.
