Help With Integration Calculation - Struggling?

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The discussion focuses on solving the integral \(\int_{-\infty}^{-\0.4088} e^{-u^2/2} du\) and clarifying the correct formulation of the integral. Dan emphasizes the necessity of including a variable in the expression and provides a method for evaluating the integral using polar coordinates. The key takeaway is the transformation of the double integral into polar coordinates, which simplifies the calculation.

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\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
 
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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
 

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