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## Homework Statement

I have the following integral I wish to solve (preferably analytically):

$$ I(x,t) = \int_{-\infty}^{0} \exp{[-(\sigma^2 + i\frac{t}{2})p^2 + (2\sigma ^2 p_a + ix)p]} \ dp$$

where ##x## ranges from ##-\infty## to ##\infty## and ##t## from ##0## to ##\infty##. ##\sigma## and ##p_a## are positive and real

letting ##(\sigma^2 + i\frac{t}{2}) = g## and ##(2\sigma ^2 p_0 + ix) = h##,

$$I(x,t) = \int_{-\infty}^{0} e^{-(gp^2 - hp)} \ dp$$

Thus far, I have tried writing this in the form of ##\text{erf}## but it becomes eye-wateringly complex (I will write my attempt below). Are there any more efficient ways to compute this integral, either numerically or analytically?

Thanks very much in advance for any assistance.

## Homework Equations

## The Attempt at a Solution

I simplify the integral by completing the square, which gives:

$$e^{\frac{h^2}{4g}} \int_{-\infty}^{0} e^{-(\sqrt{g}p \ - \frac{h}{2\sqrt{g}})^2} \ dp$$

I then change variables with ##t = (\sqrt{g}p \ - \frac{h}{2\sqrt{g}})##, giving

$$\frac{e^{\frac{h^2}{4g}}}{\sqrt{g}} \int_{z_2}^{z_1} e^{-t^2} \ dt$$

with ##z_1 = - \ \frac{h}{2\sqrt{g}}## and ## z_2 = \lim_{\ p \to -\infty } (\sqrt{g}p \ - \frac{h}{2\sqrt{g}}) ##

I then write this in term of the error function

$$I = \frac{\sqrt{\pi}}{2 } \ \frac{e^{\frac{h^2}{4g}}}{\sqrt{g}} \Big[ \text{erf} (z_1) - \text{erf} (z_2) \Big]$$

However, I am unable to determine whether or not the term ##\text{erf} (z_2)## converges or diverges, or indeed what its value is.