I need to evaluate the following integral. $$\int_{0}^{\infty} e^{-cx^{2}}\sin(tx) ~dx$$
Here's what I've done so far.
$$I(t) = \int_{0}^{\infty} e^{-cx^{2}}\sin(tx)~dx$$ Then differentiating under the integral sign I get:
$$I'(t) = \int_{0}^{\infty} xe^{-cx^{2}}\cos(tx)~dx$$
This is pretty straightforward and integrating by parts I get this:
$$I '(t) =\frac{1}{2c}+\frac{t}{2c}\int_{0}^{\infty} e^{-cx^{2}}\sin(tx)~dx$$ which can be written as
$$I'(t) = \frac{1}{2c}+\frac{t}{2c}I(t)$$
Rearranging I end up with the following differential equation:
$$I'(t)+\frac{t}{2c}I(t)=\frac{1}{2c}$$
When I try to solve this I end up with a different result than what an online calculator got:
$$-\dfrac{\sqrt{{\pi}}\mathrm{i}\mathrm{e}^{-\frac{t^2}{4c}}\operatorname{erf}\left(\frac{\mathrm{i}t}{2\sqrt{c}}\right)}{2\sqrt{c}}$$
Any tips?