Verifying that a tensor is isotropic

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    Isotropic Tensor
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I'm supposed to verify that this fourth-rank tensor is isotropic assuming cartesian coordinates: [A]_{}[/ijkl]=[δ]_{}[/kl][δ_{}[/kl]

from what I gathered being isotropic means that it stays the same no matter what the rotation is

I have no clue how to even start this problem or what I am looking at.
 
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I recommend that, instead of saying what you "gather" isotropic means, you write out the specific definition. You use the precise words of definitions in mathematics proofs.
 
I'm not even sure how to go about doing that. I am taking math methods but have been out of school for a while and just trying to relearn things and I never took a proof class before. My book does not give any definition other than the one in English but there is no math that I can find that has the definition. Is there someplace I could go to find the definition.
 
So I figured out that I need to rotate the tensor and from there show that it is the same in the new rotation. So if I have A_ijkl=(δ_ij)(δ_kl) In order to transform it I'm not really sure how to proceed.
 
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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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