Prove that $\int_{0}^{\infty} \sin(\omega x) \ e^{i \alpha x} dx=\frac{\omega}{\omega^2 - \alpha^2} + i \frac{\pi}{2} \delta(\omega-\alpha)$ Finding the FourierSinTransform of $\cos(\alpha x)$ , $\sin(\alpha x)$
Prove that :
$$\int_{0}^{\infty} \cos(\alpha x) \sin(\omega x) dx = \frac{\omega}{\omega^2 - \alpha^2}$$
And $$\int_{0}^{\infty} \sin(\alpha x) \sin(\omega x) dx = \frac{\pi}{2} \delta(\omega-\alpha)$$
I have tried to evaluate $$\int_{0}^{\infty} \sin(\omega x) \ e^{i \alpha x} dx$$
And then separating the real part and the imaginary part to get what we want 
$$\int_{0}^{\infty} \sin(\omega x) \ e^{i \alpha x} dx=\frac{-\alpha^2}{\omega^2-\alpha^2}[\frac{\sin(\omega x) e^{i \alpha x}}{i \alpha}+\frac{\omega \cos(\omega x)e^{i \alpha x}}{\alpha}]_0^\infty$$
How to get the required result ?
And If someone asked to evaluate the integral $\int_{0}^{\infty} \cos(\alpha x) \sin(\omega x) dx $ without mentioning FourierSinTransform Which is does not converge how is that make any sense ?
 A: For $ r > 0$ let 
$$T_r(\omega) = \int_0^\infty \sin(\omega x) e^{-(r-i a)x}dx =\frac{1}{2i}( \frac{1}{r-ia+i\omega}- \frac{1}{r-ia-i\omega})$$
and $T = \lim_{r \to 0} T_r$ with the limit taken in the sense of distributions, ie. $T$ is the distribution defined by 
$$\forall \phi \in C^\infty_c(\Bbb{R}) ,\qquad \langle T, \phi\rangle = \lim_{r \to 0} \ \langle T_r, \phi \rangle$$
$T_r$ is the derivative of $\frac{-1}2(\log(r -ia -i \omega)+\log(r -ia +i \omega))$ thus $T$ is the distributional derivative of $$\lim_{r \to 0} \frac{-1}2(\log(r -ia -i \omega)+\log(r -ia +i \omega))= \frac{-1}2(\log|a - \omega|+\log| a + \omega|)+ \frac{i\pi}{2} (1_{\omega > a}+1_{\omega < -a} )$$ ie. $$T =\frac12 pv(\frac{1}{a-\omega}-\frac{1}{a+\omega}) +  \frac{i\pi}{2} \delta(\omega) = pv(\frac{\omega}{a^2-\omega^2})+  \frac{i\pi}{2}(\delta(\omega-a)-\delta(\omega+a))$$
A: These kind of integrals often appear in physics and they are only definite after regularization. Specifically, this means for the integral in your title that one calculates $I:=\lim_{\eta \to 0^+}\int_{0}^{\infty} \sin (\omega x) \text{e}^{\text{i} \alpha x} \text{e}^{-\eta x}\text{d}x$. Using the exponential representation for the sinus function, $\sin(\omega x) =  \frac{1}{2\text{i}}\left( \text{e}^{\text{i} \omega x} - \text{e}^{-\text{i} \omega x}\right)$, the evaluation of the integral reads:
\begin{align}
I&=\lim_{\eta \to 0^+}\frac{1}{2\text{i}}\int_{0}^{\infty}\left(
\text{e}^{\text{i} \left(\omega + \alpha \right)x}
-\text{e}^{-\text{i}\left(\omega - \alpha \right) x}
\right)\text{e}^{-\eta x}\text{d}x\\
&=\lim_{\eta \to 0^+} \frac{1}{2\text{i}}\left(
\frac{\text{e}^{-\eta x}\text{e}^{\text{i} \left(\omega + \alpha \right)x}}{\text{i}\left(\omega + \alpha\right) - \eta}
- 
\frac{\text{e}^{-\eta x}\text{e}^{-\text{i} \left(\omega - \alpha \right)x}}{-\text{i}\left(\omega - \alpha\right) - \eta}
\right)\Bigg|_{0}^{\infty}
\\
 &=
\lim_{\eta \to 0^+} -\frac{1}{2\text{i}}\left(
\frac{1}{\text{i}\left(\omega + \alpha\right) - \eta}
+ 
\frac{1}{\text{i}\left(\omega - \alpha\right) + \eta}
\right)\\
 &=
\lim_{\eta \to 0^+} \frac{1}{2}\left(
\frac{1}{\left(\omega + \alpha\right) +\text{i} \eta}
+ 
\frac{1}{\left(\omega - \alpha\right) -\text{i} \eta}
\right)\\
 &=
\lim_{\eta \to 0^+} \frac{1}{2}\left(
\frac{\left(\omega + \alpha\right) -\text{i} \eta}{\left(\omega + \alpha\right)^2 + \eta^2}
+ 
\frac{\left(\omega - \alpha\right) +\text{i} \eta}{\left(\omega - \alpha\right)^2 + \eta^2}
\right).\\
 \end{align}
Now we reorder, 
\begin{multline}
I = \underbrace{\lim_{\eta \to 0^+} \frac{1}{2}\left( 
\frac{\omega + \alpha}{\left(\omega + \alpha\right)^2 + \eta^2}
+
\frac{\omega - \alpha }{\left(\omega - \alpha\right)^2 + \eta^2}\right)}_{\frac{\omega}{\omega^2-\alpha^2}}
-\frac{\text{i}}{2} \underbrace{\lim_{\eta \to 0^+} \frac{\eta}{\left(\omega +\alpha\right)^2 + \eta^2}}_{\pi\delta\left(\omega + \alpha\right)} + \frac{\text{i}}{2} \underbrace{\lim_{\eta \to 0^+} \frac{\eta}{\left(\omega -\alpha\right)^2 + \eta^2}}_{\pi\delta\left(\omega - \alpha\right)},
\end{multline}
which finally yields
\begin{equation}
I = \frac{\omega}{\omega^2-\alpha^2} +  \frac{\text{i}\pi}{2}\left(\delta\left(\omega - \alpha\right) - \delta\left(\omega + \alpha\right)\right).
\end{equation}
Perhaps you could double check my derivation and your supposed solution, since I get an extra $\delta\left(\omega + \alpha\right)$ term.
A: Hint
$$\sin(\omega x)=\frac{e^{i\omega x}-e^{-i\omega x}}{2i}.$$
