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Using the normalization integration for a Gaussian random variables,find an analytic expression(closed-form solution) for the following integral

$I=\int^{\infty}_{-\infty}e^{(-(ax^2+bx+c))}dx$,where $a \gt 0,b $ and $c$ are constants

There is a Hint below the question,but i don't know how to use this hint to calculate this question

Hint: Use the Gaussian integration $\frac{1}{\sigma \sqrt{2 \pi}}\int^{\infty}_{-\infty}e^{- \frac{1}{2} \frac{(x-m)^2}{\sigma ^2}}dx=1$

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You can write: $$ a x^2 +b x +c= a \left(x+\frac{b}{2a} \right)^2+c-\frac{b^2}{4a}$$ So by linearity of the integral: $$I=e^{-c+\frac{b^2}{4a}} \int_{-\infty}^{+\infty}e^{-\frac{1}{2}\frac{(x+\frac{b}{2a})^2}{1/(2a)}} dx $$ which is exactly of the form needed.

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  • $\begingroup$ may i ask how does $a(x- \frac{b}{2a})^2 +c - \frac{b^2}{4a^2}$ come?i first,i thought it is from the closed-form solution,but the closed-form solution should be $x^2 = \frac{-4ac}{(2a)^2}$ or $\frac{-2b^2+4ac}{(2a)^2}$ $\endgroup$ – XM551 Apr 10 '18 at 5:50
  • $\begingroup$ I'm ashamed but my expression was indeed incorrect, the new one should be the correct one. $\endgroup$ – Delta-u Apr 10 '18 at 6:07
  • $\begingroup$ but why is a $(x+\frac{b}{2a})+c- \frac{b^2}{4a^2}$,how is this be calculated? $\endgroup$ – XM551 Apr 10 '18 at 9:07
  • $\begingroup$ The idea is to have an expression of the form $a (x-x_0)^2+d$ as you know that using $e^{a (x-x_0)^2+d}=e^{a (x-x_0)^2}e^{d}$ the problem is a the form you need. To find $x_0$ and $d$ you make the expansion: $$a(x-x_0)^2+d=ax^2-2 a x_0 x +d-a x_0^2$$ which give $b=-2a x_0$ and $c=d-a x_0^2$. For more details see for example this article on Wikipedia. $\endgroup$ – Delta-u Apr 10 '18 at 11:51

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