Find an integral $ \int_0^{+\infty} e^{-x^2} \sinh(2ax) d x. $

Question

Find an integral $$ \int\limits_0^{+\infty} e^{-x^2} \sinh(2ax) d x. $$ Residue theorem. Or what?

Answer 1

$$ \begin{align} \int_0^\infty e^{-x^2}\sinh(2ax)\,\mathrm{d}x &=\frac12\int_0^\infty e^{-x^2}\left(e^{2ax}-e^{-2ax}\right)\,\mathrm{d}x\\ &=\frac12e^{a^2}\int_0^\infty e^{-(x-a)^2}\,\mathrm{d}x\,-\frac12e^{a^2}\int_0^\infty e^{-(x+a)^2}\,\mathrm{d}x\\ &=\frac12e^{a^2}\int_{-a}^\infty e^{-x^2}\,\mathrm{d}x\,-\frac12e^{a^2}\int_a^\infty e^{-x^2}\,\mathrm{d}x\\ &=\frac12e^{a^2}\int_{-\infty}^a e^{-x^2}\,\mathrm{d}x\,-\frac12e^{a^2}\int_{-\infty}^{-a} e^{-x^2}\,\mathrm{d}x\\ &=\frac12e^{a^2}\int_{-a}^a e^{-x^2}\,\mathrm{d}x\\ &=\frac{\sqrt\pi}2e^{a^2}\operatorname{erf}(a) \end{align} $$