Everyone has a special memory from their multivariable calc class deriving the famous Gaussian integral:
$$ \int_0^\infty e^{-x^2} \,dx = \frac{\sqrt\pi}{2}$$
A more general case is easy to find online and (not too hard to do yourself):
$$\int_{-\infty}^\infty e^{-ax^2+bx+c} \,dx = \sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}+c}$$
While I was trying to solve a certain Laplace Transform, I went looking for the answer to a similar generalization over the postive real axis:
$$\int_0^\infty e^{-ax^2+bx+c} \, dx $$
Any help approaching this last question would be extremely appreciated!
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$\begingroup$You can also express the integral in terms of the Lower incomplete Gamma function.$$\int _0^{\infty }e^{-ax^2+bx+c}\:dx=e^c\underbrace{\int _0^{\infty }e^{-a\left(\left(x-\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)}\:dx}_{u=x-\frac{b}{2a}}$$$$=e^c\int _{-\frac{b}{2a}}^{\infty }e^{-a\left(u^2-\frac{b^2}{4a^2}\right)}\:du$$$$=e^{c+\frac{b^2}{4a}}\int _{-\frac{b}{2a}}^{\infty }e^{-au^2}\:du=e^{c+\frac{b^2}{4a}}\int _0^{\infty }e^{-au^2}\:du\:-e^{c+\frac{b^2}{4a}}\underbrace{\int _0^{-\frac{b}{2a}}e^{-au^2}\:du}_{u=-u}$$$$=e^{c+\frac{b^2}{4a}}\int _0^{\infty }e^{-au^2}\:du\:+e^{c+\frac{b^2}{4a}}\underbrace{\int _0^{\frac{b}{2a}}e^{-au^2}\:du}_{t=au^2}$$$$=\frac{1}{2}\sqrt{\frac{\pi }{a}}\:e^{c+\frac{b^2}{4a}}+\frac{e^{c+\frac{b^2}{4a}}}{2\sqrt{a}}\int _0^{\frac{b^2}{4a}}e^{-t}\:t^{-\frac{1}{2}}\:dt=\frac{e^{c+\frac{b^2}{4a}}}{2\sqrt{a}}\left(\sqrt{\pi }+\gamma \left(\frac{1}{2},\frac{b^2}{4a}\right)\right)$$If you want to express it in terms of the Error function you can use the following identity$$\gamma \left(\frac{1}{2},x\right)=\sqrt{\pi }\:\text{erf}\left(\sqrt{x}\right)$$
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