# $\int_{-\infty}^{\infty}i\cdot \sin(x)\sin(2{\pi}kx)\;dx$ during Fourier transform

I am trying to do a time-to-frequency domain transform using Fourier transform. My function is very simple: $$f(x) = \sin(x)$$ By definition its Fourier transform should be: $$F(k) = \int_{-\infty}^{\infty}\sin(x)e^{-2{\pi}ikx}\;dx \\ = \int_{-\infty}^{\infty}\sin(x)\left[\cos(2{\pi}kx)-i\cdot\sin(2{\pi}kx)\right]dx \\ = \int_{-\infty}^{\infty}\sin(x)\cos(2{\pi}kx)dx - \int_{-\infty}^{\infty}i\cdot\sin(x)\sin(2{\pi}kx)dx \\ = 0 - \int_{-\infty}^{\infty}i\cdot\sin(x)\sin(2{\pi}kx)dx$$ Am I right so far? What's next? I couldn't calculate the term $\int_{-\infty}^{\infty}i\cdot\sin(x)\sin(2{\pi}kx)dx$... I tried integration by parts but I still couldn't get it.

-

The original integral doesn't converge. The Fourier transform $\mathscr F sin (k)$ gives us the coefficient of the basis element $e{^{i2\pi kx}}$.