I don't understand why the following is wrong: link here
Should dθ sinθ be converted to -dcosθ? If yes should dθ sinθ be converted to -dcosθ for every integral involving cosθ?
Integrating δ(1 - cosθ) over the entire solid angle in spherical coordinates.
0
$\begingroup$
integration
dirac-delta
solid-angle
-
0Are you saying that the answer $0$, which is derived in the image, is wrong? – 2017-02-17
-
0Perhaps I should ask if it is wrong... Is it? And what happens if we replace dθ sinθ with -dcosθ? – 2017-02-17
-
1The answer is indeed zero, using that change of variables isn't necessary to see why it is zero. – 2017-02-17
-
0It's a small idea to u substitution. The bounds will instead be between $-1$ and $1$ – 2017-02-17
-
0So do I treat the delta function like any other function taking the difference of the evaluates of the function for the two boundaries? I thought that integrating the delta function over a range that includes the values for which the nested function becomes zero always yields 1. – 2017-02-17
1 Answers
0
The answer is $0$. If you want to use $-d \cos(\theta)$ in place of $d\theta \sin \theta$, then you get $$ C \int_0^{2\pi} d\phi \int_{-1}^1 \delta_{1-t} dt, $$ which is $0$ if you define $\delta$ function appropriately.