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I want to compute the following integral

$$\int\limits_{-\infty}^t e^{-\frac{1}{2 a}(c-x)^2} dx$$

I substitute $y=\frac{c-x}{\sqrt{a}}$. Thus I have $dy=-\frac{dx}{\sqrt{a}}$ and the upper limit $\frac{c-t}{\sqrt{a}}$. Then I have

$$-\sqrt{a}\int\limits_{-\infty}^{\frac{c-t}{\sqrt{a}}} e^{-y^2} dy=\sqrt{a}\int\limits^{\infty}_{\frac{c-t}{\sqrt{a}}} e^{-y^2} dy=\sqrt{2\pi} \sqrt{a}(1-\Phi(\frac{c-t}{\sqrt{a}}))$$

Is this correct?Can I simplify this further?

Yuriy S
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    You lost the $2$ from the beginning. Set $y=\frac{c-x}{\sqrt{2a}}$. You cannot simplify further. – Claude Leibovici Feb 20 '15 at 11:14
  • You are very welcome ! May I suggest you correct the post and put the correct results ? By the way, I am not sure that $\Phi$ is the normal way to represent the error function. Cheers :-) – Claude Leibovici Feb 21 '15 at 04:17
  • No, but $\Phi$ is the normal (haha) way to write the cdf of the standard normal distribution. – GEdgar Aug 06 '16 at 17:43

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