Prove that

a) $\displaystyle \int_0^{\infty} e^{-\xi^2} \frac{\sin 2 \xi y}{\xi} d \xi = \frac{\pi}{2} erf y$

b) $\displaystyle \int_0^{\infty} e^{-\xi^2} \sin 2 \xi y} d \xi = \frac{\sqrt{\pi}}{2} e^{-y^2} erf y$

where $\displaystyle erf x = \frac{2}{\sqrt{\pi}} \int_0^x e^{-\xi^2} d \xi $

I tried out integration by parts but was ending up in indefinite form. I think it might need a smart substitution but could not figure it out any help is appreciated.

Regards,

~Kalyan.