In the post, there are 4 methods to find the integral
$$ \int_0^{\infty} e^{-a x^2} \cosh (b x) d x= \frac{e^{\frac{b^2}{4 a}}}{2} \sqrt{\frac{\pi}{a}}, $$ I believe that we can evaluate its partner integral $$\int_0^{\infty} e^{-a x^2} \sinh (b x) d x$$
in a similar manner.
However, the integrand is unfortunately odd and needs two results involving the error function $\operatorname{erf}(x)$ and its complementary function $\operatorname{erfc}(x)$: $$ \int_0^{\infty} e^{-h(x-k)^2} d x=\frac{1}{2}\sqrt{\frac{\pi}{h}} \left[\operatorname{erf}(\sqrt{h} k)+1\right] \tag*{(1)} $$ and $$ \int_0^{\infty} e^{-h(x+k)^2} d x=\frac{1}{2} \sqrt{\frac{\pi}{h}} \operatorname{erfc}(\sqrt{h} k) \tag*{(2)} $$
where $h,k>0$ and $ \operatorname{erfc}(x)=1-\operatorname{erf}(x)\cdots (3). $
Equations $(1)$ and $(2)$ can be proved by the substitutions $y=\sqrt h(x+k)$.
Using the definition of $\cosh x$ and grouping the indices by the method of completing squares, we have $$ I=\int_0^{\infty} e^{-a x^2} \sinh(b x) d x=\frac{e^{\frac{b^2}{4 a}}}{2}\left.\int_{-\infty}^{\infty}\left(e^{-a\left(x-\frac{b}{2 a}\right)^2}-e^{-a\left(x+\frac{b}{2 a}\right)^2}\right) d x\right. $$
Now using $(1)-(3)$ yields $$ \begin{aligned} I & =\frac{e^{\frac{b^2}{4 a}}}{2}\left[\frac{1}{2}\left(\operatorname{erf}\left(\frac{b}{2 \sqrt{a}}+1\right)-\frac{1}{2} \operatorname{erfc}\left(\frac{b}{2 \sqrt{a}}\right)\right]\right. =\frac{e^{\frac{b^2}{4 a}}}{2} \sqrt{\frac{\pi}{a}} \operatorname{erf}\left(\frac{b}{2 \sqrt{a}}\right) \end{aligned} $$
Are there any other forms of answers?
Alternative methods and comments are highly appreciated.
Feynman’s Trick
Considering the integral parametrized by $b$ $$ I(b)=\int_0^{\infty} e^{-ax^2} \sinh (b x) d x $$ with $I(0)=0$.
Differentiating w.r.t. $b$ gives $$ \begin{aligned} I^{\prime}(b) & =\int_0^{\infty} x e^{-ax^2} \cosh (b x) d x \\ & =-\frac{1}{2a} \int_0^{\infty} \cosh (b x) d\left(e^{-ax^2}\right) \\ & =-\frac{1}{2a}\left(\left[e^{-ax^2} \cosh (b x)\right]_0^{\infty}+ \int_0^{\infty} e^{-ax^2} \sinh (bx) d x\right)\\&=\frac{1}{2a}+\frac{b}{2a} I(b) \end{aligned} $$
Rearranging and solving by integrating factor gives $$ \begin{aligned} & I^{\prime}(b)-\frac{b}{2 a} I(b)=\frac{1}{2 a} \\ \Leftrightarrow \quad & {\left[e^{-\frac{b^2}{4 a}} I(b)\right]^{\prime}=\frac{1}{2 a} e^{-\frac{b^2}{4 a}} } \\ \Leftrightarrow \quad & e^{-\frac{b^2}{4 a}} I(b)=\frac{1}{2 } \sqrt{\frac{\pi}{a}} \operatorname{erf}\left(\frac{b}{2 \sqrt{a}}\right) \\ \Leftrightarrow \quad & I(b)= \frac{e^{\frac{b^2}{4 a}}}{2} \sqrt{\frac{\pi}{a}} \operatorname{erf}\left(\frac{b}{2 \sqrt{a}}\right) \end{aligned} $$