Lorentz invariance of an equation (metric)

felixphysics
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I'm not sure what you mean by "metric is invariant." The metric transforms like a tensor, covariant or contravariant depending on whether the indices are up or down. The line element ##ds^2 = g_{\mu\nu} dx^\mu dx^\nu## is Lorentz invariant, but that's not what appears in your formula.

You should write down the explict transformation rules for ##V^\mu##, ##\partial/\partial x^\mu##, ##g^{\nu\sigma}##, and ##g_{\nu\sigma}##. You will need to combine them in your expression and carefully apply the derivatives with the appropriate product rules for differentiation.
 
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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