Writing integrals in terms of the error function

Click For Summary

Homework Help Overview

The discussion revolves around expressing a specific integral involving an exponential function in terms of the error function. The integral in question is defined with limits from negative infinity to zero, and participants are exploring how to manipulate it to fit the form of the error function.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various change of variables to transform the integral into a suitable form for the error function. There are attempts to rewrite the integral and questions about the correctness of these transformations.

Discussion Status

Several participants have provided guidance on rewriting the integral and have suggested exploring the definition of the error function. There is an ongoing examination of the limits and the implications of different variable substitutions, with no clear consensus yet on the best approach.

Contextual Notes

Participants note the need to express the integral with limits that align with the error function's definition, which has led to discussions about potential errors in rewriting the integral. There is also mention of formatting issues in LaTeX when expressing the error function.

WWCY
Messages
476
Reaction score
15

Homework Statement


I have the following integral,

$$\frac{1}{\sigma \sqrt{2\pi} t} \int_{-\infty}^{0} \exp[\frac{-1}{2\sigma ^2} (\frac{x-x_0}{t} - p_0)^2]dx$$
that I wish to write in terms of the error function,
$$erf(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-g^2}dg$$
However, I can't seem to make my limits fit that of ##erf(x)## despite trying a change of variables like letting ## g = \frac{-1}{\sqrt{2\sigma ^2}} (\frac{x-x_0}{t} - p_0)##

This is my first time dealing with such a function, and pointers are greatly appreciated.

Homework Equations

The Attempt at a Solution

 
Physics news on Phys.org
WWCY said:

Homework Statement


I have the following integral,

$$\frac{1}{\sigma \sqrt{2\pi} t} \int_{-\infty}^{0} \exp[\frac{-1}{2\sigma ^2} (\frac{x-x_0}{t} - p_0)^2]dx$$
that I wish to write in terms of the error function,
$$erf(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-g^2}dg$$
However, I can't seem to make my limits fit that of ##erf(x)## despite trying a change of variables like letting ## g = \frac{-1}{\sqrt{2\sigma ^2}} (\frac{x-x_0}{t} - p_0)##

This is my first time dealing with such a function, and pointers are greatly appreciated.

Homework Equations

The Attempt at a Solution

PF rules really do require you to tell us more. Show your actual work; where does it fail?

For constants ##a,b,c,k,w## (with ##c > 0##), can you express
$$\int_a^b k e^{-c(x-w)^2} \, dx$$
in terms of ##\text{erf}( \cdot)##?
 
Thanks for the response

Ray Vickson said:
PF rules really do require you to tell us more. Show your actual work; where does it fail?
For constants ##a,b,c,k,w## (with ##c > 0##), can you express
$$\int_a^b k e^{-c(x-w)^2} \, dx$$
in terms of ##\text{erf}( \cdot)##?

Using the integral given as an example, I tried a change of variables ##f = \sqrt{c}(x - w)^2##,
which led to
$$\frac{k}{\sqrt{c}} \int_{f_1}^{f_2}e^{-f^2}df $$
I can't tell if this was the right "first-step" to take.

May I know what are the concepts in play? For example, is this just an issue of changing variables, or is there more to it?
 
Rewrite the integral as two integrals from ##_-\infty## to some upper limit.
 
Thanks for the response

Orodruin said:
Rewrite the integral as two integrals from ##_-\infty## to some upper limit.

Do you mean something like this?
$$\frac{k}{\sqrt{c}} (\int_{-\infty}^{f_2}e^{-f^2}df - \int_{-\infty}^{f_1}e^{-f^2}df)$$

If so, how do I continue? Thank you for assisting.
 
Sorry, I meant from ##0## to some upper limit ... :rolleyes:

Edit: Also, your rewriting is not correct. There is a minor error in it. Try to find it.
 
Orodruin said:
Sorry, I meant from ##0## to some upper limit ... :rolleyes:

Edit: Also, your rewriting is not correct. There is a minor error in it. Try to find it.

Oops, I believe I have spotted it.

$$\frac{k}{\sqrt{c}} (\int_{0}^{f_2}e^{-f^2}df - \int_{0}^{f_1}e^{-f^2}df)$$
where ##f_1 = \sqrt{c} (a - w), f_2 = \sqrt{c} (b - w)##

Is this correct? And how do I proceed?

Thanks!
 
Uhhhmm... Apply the definition of the error function?
 
I got,
$$\frac{k}{2} \sqrt{\frac{\pi}{c}}[erf(f_2) - erf(f_1)]$$

Orodruin said:
Uhhhmm... Apply the definition of the error function?

Apologies, I meant to ask how I would fit in the limits of ##-\infty## and ##0## as per my original question.

For ##b = 0, a = -\infty##,
$$\frac{k}{2} \sqrt{\frac{\pi}{c}}[erf(- \sqrt{c} w) - \frac{2}{\sqrt{\pi}} \int_{0}^{-\infty}e^{-f^2}df]$$
$$\frac{k}{2} \sqrt{\frac{\pi}{c}}[erf(- \sqrt{c} w) + 1]$$

Am I doing it right? Thank you!
 
  • #10
WWCY said:
I got,
$$\frac{k}{2} \sqrt{\frac{\pi}{c}}[erf(f_2) - erf(f_1)]$$
Apologies, I meant to ask how I would fit in the limits of ##-\infty## and ##0## as per my original question.

For ##b = 0, a = -\infty##,
$$\frac{k}{2} \sqrt{\frac{\pi}{c}}[erf(- \sqrt{c} w) - \frac{2}{\sqrt{\pi}} \int_{0}^{-\infty}e^{-f^2}df]$$
$$\frac{k}{2} \sqrt{\frac{\pi}{c}}[erf(- \sqrt{c} w) + 1]$$

Am I doing it right? Thank you!

Yes, except that you are not writing "erf" correctly in LaTeX. You should say "##\text{erf}(f_2)##" instead of "##erf(f_2)##". Right-click on the expressions to see the difference.
 
  • #11
Ray Vickson said:
Yes, except that you are not writing "erf" correctly in LaTeX. You should say "##\text{erf}(f_2)##" instead of "##erf(f_2)##". Right-click on the expressions to see the difference.

Thank you, I'll keep that in mind!
 
  • #12
WWCY said:
Thank you, I'll keep that in mind!

Also: for most of the common functions (sin, cos, tan, arcsin,arccos, arcsin,cot,ln,log,max,min,lim,exp) and several others, it is enough to just put a "\" in front, so you get ##\sin \theta## (which looks good) instead of ##sin \theta## (which looks ugly and is hard to read).
 
  • Like
Likes   Reactions: WWCY
  • #13
Ray Vickson said:
Also: for most of the common functions (sin, cos, tan, arcsin,arccos, arcsin,cot,ln,log,max,min,lim,exp) and several others, it is enough to just put a "\" in front, so you get ##\sin \theta## (which looks good) instead of ##sin \theta## (which looks ugly and is hard to read).
When the command does not have a "\" form, we need to manually force a non-math-italic font by putting the command inside a "\text{ }" construct, so we get ##P(2\; \text{heads}) = 1/4## instead of ##P(2\; heads) = 1/4##. That is why we needed to say \text{erf}(x): the function "erf" is not on the list of commands/functions having a short "\" form.
 
Last edited:
  • Like
Likes   Reactions: WWCY
  • #14
Actually, the \text command does not really do the same thing as the typical function commands. The spacing will be off. The appropriate way of putting new function commands into LaTeX is to use the \DeclareMathOperator command. If you just need it once, you should use \operatorname, i.e., ##\operatorname{erf}(x)##. Compare the \operatorname solution to "a erf(x)" with the \text solution:
$$
a \operatorname{erf}(x) \quad \mbox{vs} \quad a \text{erf}(x)
$$
 
  • Like
Likes   Reactions: WWCY
  • #15
I have attempted working out my initial problem in the same manner as above, could someone give the working a look? Thanks in advance!
$$I = \frac{1}{\sigma \sqrt{2\pi} t} \int_{-\infty}^{0} \exp[\frac{-1}{2\sigma ^2} (\frac{x-x_0}{t} - p_0)^2]dx$$
Let ##g = \frac{1}{\sigma \sqrt{2}} ( \frac{x - x_0}{t} -p_0 ) ## and ##dg = \frac{1}{\sigma t \sqrt{2}} dx##
$$I = \frac{1}{\sqrt{\pi}} \int_{g_2}^{g_1} e^{-g^2}dg$$
$$I = \frac{1}{\sqrt{\pi}} [ \int_{0}^{g_2} e^{-g^2}dg - \int_{0}^{g_1} e^{-g^2}dg]$$
$$I = \frac{1}{\sqrt{\pi}} \frac{\sqrt{\pi}}{2}[\operatorname{erf(g_2)} - \frac{2}{\sqrt{\pi}} \int_{0}^{-\infty} e^{-g^2}dg ]$$
$$I = \frac{1}{2} [ \operatorname{erf}(\frac{1}{\sigma \sqrt{2}} [-\frac{x_0}{t} - p_0] ) + 1]$$
$$I = \frac{1}{2} [ \operatorname{erf}(-\frac{1}{\sigma \sqrt{2}} [\frac{x_0}{t} + p_0] ) + 1]$$
 
  • #16
Orodruin said:
Actually, the \text command does not really do the same thing as the typical function commands. The spacing will be off. The appropriate way of putting new function commands into LaTeX is to use the \DeclareMathOperator command. If you just need it once, you should use \operatorname, i.e., ##\operatorname{erf}(x)##. Compare the \operatorname solution to "a erf(x)" with the \text solution:
$$
a \operatorname{erf}(x) \quad \mbox{vs} \quad a \text{erf}(x)
$$

Thanks: I did not know that. However, I would rather type " a \: \text{erf}(x)" than "a \operatorname{erf}(x)". I have long been aware of that spacing problem and have used spacers to compensate.

Now that I know about "\operatorname" I can see defining a short form of it in the document preamble.
 
Last edited:
  • #17
Ray Vickson said:
Thanks: I did not know that. However, I would rather type " a \: \text{erf}(x)" than "a \operatorname{erf}(x)". I have long been aware of that spacing problem and have used spacers to compensate.

Now that I know about "\operatorname" I can see defining a short form of it in the document preamble.
If you are going to use a single function often, I strongly recommend \DeclareMathOperator instead.
 
  • #18
Could someone assist regarding post 15? Many thanks.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K