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?
Best Answer
$$ \begin{align} \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin (\alpha) \cos (x \alpha)}{\alpha} \ d \alpha &= \frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \big((1+x) \alpha \big)+\sin \big( (1-x) \alpha \big)}{\alpha} \ d \alpha \\ &= \frac{1}{\pi} \Big(\text{sgn}(1+x) \frac{\pi}{2}+\text{sgn}(1-x) \frac{\pi}{2} \Big) \\ &= \frac{1}{2} \Big(\text{sgn}(1+x) + \text{sgn}(1-x) \Big) \\ &= \begin{cases} \frac{1}{2}(-1+1) = 0 & \text{if} \ x <-1 \\ \frac{1}{2}(0+1) = \frac{1}{2} & \text{if} \ x = -1 \\ \frac{1}{2} (1+1) = 1 & \text{if} -1 < x <1 \\ \frac{1}{2}(1+0) = \frac{1}{2} & \text{if} \ x = 1 \\ \frac{1}{2} (1-1) = 0 & \text{if} \ x >1\end{cases} \end{align}$$