Definite Integral of exponential function with changed limits

Question

I want to integrate in terms of x:

I know the integral of:

but can't seem to figure out the first integral.

Answer 1

$$ \mathrm{e}^{-ax^2-bx} = \mathrm{e}^{-a\left(\left(x+\frac{b}{2a}\right)^2 -\frac{b^2}{4a^2}\right)} $$ So we have an integral $$ \int_{0}^{\infty}\mathrm{e}^{-a\left(\left(x+\frac{b}{2a}\right)^2 -\frac{b^2}{4a^2}\right)} dx = \mathrm{e}^{\frac{b^2}{4a}}\int_{0}^{\infty}\mathrm{e}^{-a\left(x+\frac{b}{2a}\right)^2} dx $$ using $z^2 = a\left(x+\frac{b}{2a}\right)^2 \to dz = \sqrt{a} dx$ so we have $$ \frac{\mathrm{e}^{\frac{b^2}{4a}}}{\sqrt{a}}\int_{\frac{b}{\sqrt{4a}}}^{\infty}\mathrm{e}^{-z^2}dx $$ using $$ 1-\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{x}^{\infty}\mathrm{e}^{-z^2}dx $$ so we get $$ \int_{0}^{\infty} \mathrm{e}^{-ax^2-bx} dx = \frac{\sqrt{\pi}}{2}\frac{\mathrm{exp}\left(\frac{b^2}{4a}\right)}{\sqrt{a}}\left(1-\mathrm{erf}\left(\frac{b}{\sqrt{4a}}\right)\right) $$