Why integral $I(b,x) = \int\limits_x^\infty {{e^{ - ({y^2} + by)}}dy = \sqrt \pi {e^{{b^2}/4}}N( - \sqrt 2 - b/\sqrt 2 )} $ where the function $N$ is defined as $N(x)=\frac{1}{{\sqrt {2\pi } }}\int\limits_{ - x}^\infty {{e^{ - {y^2}/2}}dy = } \frac{1}{{\sqrt {2\pi } }}\int\limits_{ - \infty }^x {{e^{ - {y^2}/2}}dy} $
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Hint: complete the square with $y^2+by$ and then make a change of variables to turn the nonconstant part into $u^2/2$. – Greg Martin Mar 29 '21 at 07:08
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@GregMartin Could you give me a bit more details, please – Adam Mar 29 '21 at 07:31
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Consider the more general case $$I= \int e^{-(a y^2+b y+c)}\,dy $$ As siad in comments, complete the square $$ay^2+by+c=a \left(y+\frac{b}{2 a}\right)^2-\frac{b^2-4ac}{4a}$$ $$I=\exp\Bigg[\frac{b^2-4ac}{4a} \Bigg]\int \exp\Bigg[-a \left(y+\frac{b}{2 a}\right)^2\Bigg]\,dy$$ Let $$y=-\frac{b}{2 a}-\frac{t}{\sqrt{2} \sqrt{a}}\implies dy=-\frac{dt}{ \sqrt{2a}}$$ $$\int \exp\Bigg[-a \left(y+\frac{b}{2 a}\right)^2\Bigg]\,dy=-\frac{1}{ \sqrt{2a}}\int e^{-\frac{t^2}{2}}\,dt$$ Recombine everything to get $I$
Claude Leibovici
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