Numerical/Analytical Solution to a Complex Integral

[WWCY]

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.

 

[Ray Vickson]

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.

I think you should be able to prove the following:
(1) If $f(N) = \text{erf}(aN + ib)$ with $a,b$ constant, then $\lim f(N)$ exists and is finite as $N \to \pm \infty$
(2) If $f(N) = \text{erf}(a + i b N)$ for constants $a,b$, then $b \neq 0 \Rightarrow f(N)$ diverges as $N \to \pm \infty$. You ought to be able to do this by looking at the integral and seeing whether it can converge or not. So, if the imaginary part of the erf function stays bounded but the real part becomes infinite, you get a finite answer; if the real part remains bounded while the imaginary part becomes infinite you will have an infinite answer.

A trickier case (like yours) is where the real and imaginary parts go to $\infty$ together, such as for $f(N) = \text{erf}((a+ib)N)$ with real $a,b$. I looked at this in Maple, and think that for $a,b > 0$ at least we have that $f(N)$ converges if $b \leq a$ but diverges if $b > a$. The attached file is from a Maple session in which I looked at $f_1(x) = \text{erf}((1+i)x), f_2(x)= \text{erf}((1+1.01) x)$ and $f_3(x) = \text{erf}((1+0.9 i)x))$, giving plots of the real and imaginary parts from $x=0$ to some moderately large positive value. The graphs show quite convincingly that we get convergence for $f_1$ and $f_3$ but divergence for $f_2.$

The challenge is to somehow prove all this "analytically", but to date I don't know how to do it.

 

[WWCY]

Thank you for the reply!

A trickier case (like yours) is where the real and imaginary parts go to $\infty$ together, such as for $f(N) = \text{erf}((a+ib)N)$ with real $a,b$. I looked at this in Maple, and think that for $a,b > 0$ at least we have that $f(N)$ converges if $b \leq a$ but diverges if $b > a$. The attached file is from a Maple session in which I looked at $f_1(x) = \text{erf}((1+i)x), f_2(x)= \text{erf}((1+1.01) x)$ and $f_3(x) = \text{erf}((1+0.9 i)x))$, giving plots of the real and imaginary parts from $x=0$ to some moderately large positive value. The graphs show quite convincingly that we get convergence for $f_1$ and $f_3$ but divergence for $f_2.$
The challenge is to somehow prove all this "analytically", but to date I don't know how to do it.

I have a few follow-up questions I'd like to ask:
1. What do the analytical expressions for the real and imaginary parts of $\text{erf} (x + iy)$ look like? I've looked around for such expressions but have been unable to find them.
2. I assume the case mentioned in the quote refers to $\text{erf} (z_2)$. My main concern in this case is that $t$ is a temporal variable, and $x$ is spatial. I'm concerned that the function diverges when these two head to infinity. Could I investigate by plotting against $x$ and $t$ (for Re and I am of erf) in Maple as you did?

I was also thinking about the following to see if we can sort this out analytically by rewriting the terms in the erf's using roots of unity. However I don't know whether or not simplifying the input actually does anything to make the problem simpler or clearer. Do you mind taking a look?

Take for instance $z_2$ where $p$ implicitly heads to $-\infty$ while $x$ goes to $\infty$
Start with
$$z_2 = \sqrt{ \sigma^2 + i\frac{t}{2} } p \ - \frac{2\sigma ^2 p_0 + ix}{2\sqrt{\sigma^2 + i\frac{t}{2} }}$$
as $t$ tends to $\infty$, the term $$\sqrt{\sigma^2 + i\frac{t}{2} } = \sqrt{e^{i\frac{\pi}{2}}} = e^{i\frac{\pi}{4}}$$
where I take the positive root of unity. I actually don't know whether or not this is a valid choice, though what I learned was that both choices are equally valid. I then rewrite $z_2$ as

$$z_2 = \frac{1}{\sqrt{2}} (1+i)p - \frac{ (2\sigma^2 p_0 + ix)(1-i) }{\frac{2}{\sqrt{2}} (1+i)(1-i)} $$

which then reduces to

$$z_2 = [\frac{1}{\sqrt{2}} (1+i)]p - \frac{1}{2 \sqrt{2}} [x + 2\sigma^2 p_0 + (x - 2\sigma^2 p_0)i ]$$

also, this means

$$z_1 = \frac{1}{2 \sqrt{2}} [x + 2\sigma^2 p_0 + (x - 2\sigma^2 p_0)i$$

Does this do anything for the problem or should I stick to the numerical methods?

Thank you again for your assistance. Apologies for my lack of understanding wrt some of these ideas.