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Evaluate $\int_0^{\infty } e^{-2 a x-x^2} dx$ without using error function

MathFC
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  • This is going to be interesting (for me, at least): without the error function... – DonAntonio Dec 02 '16 at 17:06
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    There's no elementary closed form for this integral if $a\neq 0$. One could write down an infinite series expansion, but this is the same as what the error function would give you. – Semiclassical Dec 02 '16 at 19:07

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Why so adamant about no error function? Completing the square in the exponential you have $$ \begin{align} \int_0^{\infty} e^{-x^2 - 2ax} dx = \sqrt{\pi}e^{a^2}\int_0^{\infty}\frac{1}{\sqrt{\pi}}e^{-(x+a)^2}dx. \end{align} $$ So the second term is just the Gaussian with location $-a$ and scale $1$, so there really isn't any closed form and the question just reduces to deriving approximations on the tails of Gaussians, which is another question entirely but see something like https://arxiv.org/abs/1012.2063

Nadiels
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