Integral in spherical coordinates

aaaa202
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I recently had to do an integral like the one in the thread below:
http://math.stackexchange.com/quest...-of-radial-function-without-bessel-and-neuman
The problem I had was also evaluating the product and I am quite sure that the answer in the thread is the one I need. I just don't understand it fully. They say we fix our x-vector such that its angle with the z-axis is the same as its dot product with the other vector. But isn't x an everchanging vector? I meant we are integrating over it. What am I missing?
 
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The integral will produce a function of vector ξ. For any given ξ, because of the spherical symmetry elsewhere, you can rotate your coordinates to align ξ with one of the axes. In this case, z was chosen.
 
Ahh I get it now. Though I don't see which other axis you could align it with? The inclination angle does not rotate around the x or y axis. The special reason we can use the z-axis is that the inclination azimuthal angle rotates around it. Do you not agree?
 
aaaa202 said:
Ahh I get it now. Though I don't see which other axis you could align it with? The inclination angle does not rotate around the x or y axis. The special reason we can use the z-axis is that the inclination azimuthal angle rotates around it. Do you not agree?

In principle you could choose the ξ direction for any of the axes, but maybe only choosing it as the z-axis makes the integral amenable.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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