@ssd, if you mean that the end result of the integral in the original post is half the result of the other one given in there, you are making a calculation error.
OK yf920, let's calculate the result given. The argument of the exponential function can be written as:
-ax^2-bx-c=-a\left(x+\frac{b}{2a}\right)^2 +\frac{b^2-4ac}{4a}
Putting this into the integral gives then:
I=e^{\frac{b^2-4ac}{4a}}\int_{-\infty}^{\infty} e^{-a\left(x+\frac{b}{2a}\right)^2}dx
Putting as ssd said:
\sqrt{a}\left(x+\frac{b}{2a}\right)=t
gives now:
I=\frac{e^{\frac{b^2-4ac}{4a}}}{\sqrt{a}}\int_{-\infty}^{\infty} e^{-t^2}dt
The following integral is known:
\int_{-\infty}^{\infty} e^{-t^2}dt=\sqrt{\pi}
Giving the final result as:
I=\sqrt{\frac{\pi}{a}} e^{\frac{b^2-4ac}{4a}}
Now, the original question was:
I=\int_{0}^{\infty} e^{-\left(ax^2+bx+c\right)}dx
Which is not the double of the one just derived. Just go through all the steps with this new integral limits and use the result:
erf(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{x}e^{-z^2}dz
Do these calculations, and come back with the result you found, it will be our pleasure to check it.
