Help With Integration Calculation - Struggling?

  • Context: MHB 
  • Thread starter Thread starter JWelford
  • Start date Start date
  • Tags Tags
    Integration
Click For Summary

Discussion Overview

The discussion revolves around a specific integration calculation involving the exponential function. Participants are exploring the correct formulation of the integral and addressing potential errors in the expression provided by the original poster. The conversation includes technical reasoning and mathematical exploration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • The original poster expresses uncertainty about the formulation of the integral, specifically questioning whether a variable is needed in the expression.
  • Dan suggests that there should be a variable in the integral and later corrects the expression to include an exponential term with a variable.
  • Dan introduces a related integral, I, and discusses its properties, including the use of dummy variables and the application of polar coordinates to evaluate the integral.
  • Dan provides a detailed breakdown of the steps involved in transforming the integral into a double integral in polar coordinates, leaving the final steps for others to complete.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct formulation of the integral, and multiple viewpoints regarding the necessary components of the integral remain present.

Contextual Notes

The discussion includes potential missing assumptions regarding the bounds of integration and the specific variables involved in the integral. There are also unresolved mathematical steps in the transformation process described by Dan.

Who May Find This Useful

Readers interested in integration techniques, particularly those involving exponential functions and polar coordinates, may find this discussion relevant.

JWelford
Messages
3
Reaction score
0
\int_{-0.4088}^{-\infty}\,e^{-1/2.4^2}d struggling to solve this calculation. Not sure if i have written the formula in the right way. First post on this site. thanks
 
Physics news on Phys.org
JWelford said:
\int_{-0.4088}^{-\infty}\,e^{-1/2.4^2}d struggling to solve this calculation. Not sure if i have written the formula in the right way. First post on this site. thanks
[math]\int_{-.4088}^{\infty} e^{-1/2.4} d[/math]

There needs to be a variable in there somewhere!

-Dan
 
topsquark said:
[math]\int_{-.4088}^{\infty} e^{-1/2.4} d[/math]

There needs to be a variable in there somewhere!

-Dan

woops its e^-1/2 . u^2 du

and the lower bound is minus infinity
 
JWelford said:
woops its e^-1/2 . u^2 du

and the lower bound is minus infinity
Here's a trick to remember. Let
[math]I = \int_{-\infty}^{\infty} e^{-u^2/2}~du[/math]

Now, u is a "dummy variable" so we can just as easily say that [math]I = \int_{-\infty}^{\infty} e^{-v^2/2}~dv[/math].

Since u and v are unrelated we can multiply these together:
[math]I^2 = \left ( \int_{-\infty}^{\infty} e^{-u^2/2}~du \right ) \left ( \int_{-\infty}^{\infty} e^{-v^2/2}~dv \right ) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-u^2/2} ~ e^{-v^2/2}~dv~du[/math]

So
[math]I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{(-1/2)(u^2 + v^2)} ~dv~du[/math]

Define a set of polar coordinates [math]( r, \theta )[/math] such that [math]u = r~cos( \theta )[/math] and [math]v = r~sin( \theta )[/math]. The Jacobian is equal to r and [math]u^2 + v^2 = r^2[/math], so
[math]I^2 = \int_{0}^{\infty} \int_{0}^{2 \pi} e^{-r^2/2} ~r~d \theta~dr[/math]

The rest I leave to you.

-Dan
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K