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 ?

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. $$T =\frac12 pv(\frac{1}{a-\omega}-\frac{1}{a+\omega}) + \frac{i\pi}{2} \delta(\omega) = pv(\frac{\omega}{a^2-\omega^2})+ \frac{i\pi}{2}(\delta(\omega-a)-\delta(\omega+a))$$
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 $$$$I = \frac{\omega}{\omega^2-\alpha^2} + \frac{\text{i}\pi}{2}\left(\delta\left(\omega - \alpha\right) - \delta\left(\omega + \alpha\right)\right).$$$$ Perhaps you could double check my derivation and your supposed solution, since I get an extra $$\delta\left(\omega + \alpha\right)$$ term.
• yes of course you are right. My primitive/derivative approach is quite general, for example it works to show $\lim_{A \to \infty}\int_{-A}^A e^{i xy}dy = 2\pi \delta(x)$, its major problem is we need to apply the distribution to non smooth test functions, in which case your method is better – reuns Apr 8 '19 at 5:36
Hint $$\sin(\omega x)=\frac{e^{i\omega x}-e^{-i\omega x}}{2i}.$$