# Why does $\sin{\alpha}\cdot i\sin{\alpha x}$ disappear from this integral?

In a section on fourier transforms, my textbook contains these steps for an example:

$$f(x) = \int_{-\infty}^\infty \frac{\sin{\alpha}}{\pi \alpha}e^{i\alpha x}d\alpha$$ $$= \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\sin{\alpha}(\cos{\alpha x} + i \sin{\alpha x})}{\alpha}d\alpha$$ $$= \frac{2}{\pi}\int_{0}^{\infty}\frac{\sin{\alpha}\cos{\alpha x}}{\alpha}d\alpha$$

I understand why the coefficient changes to $\frac{2}{\pi}$ and the integral changes to $\int_0^\infty$, but why does the $\sin{\alpha}\cdot i\sin{\alpha x}$ disappear in the final step? Am I missing something obvious here?

• Fourier coefficients of an odd function... – DonAntonio Feb 17 '13 at 5:02

$\dfrac1\alpha\sin\alpha\cdot i\sin\alpha x$ is an odd function of $\alpha$.
• Yes, quite the odd function, it swallowed $i$ whole! :-) – Asaf Karagila Feb 17 '13 at 5:03