A very weird improper integral on ^n

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Homework Statement


Verify that \int_{ℝ^n}exp(-\frac{λ}{2} \langle Ax, x \rangle-i \langle x,ζ \rangle )dx=(\frac{2\pi}{λ})^{\frac{1}{2}}(detA)^{-\frac{1}{2}}exp(-\frac{1}{2λ} \langle A^{-1}ζ, ζ \rangle ) where A is a symmetric matrix of complex numbers and <ReA x, x> is positive definite, and λ is a positive constant. ζ is a vector in ℝ^n

Homework Equations


Fubini's Theorem?

The Attempt at a Solution


The question is a lot easier if A is brought to diagonal form, so it is reasonable to make a change of variable that x= C y where C belongs to SO(n) such that C^-1 A C = B is diagonal. Since this change of variables means only geometrically a rotation of the R^n plane it should not change the range of values for integrating(still from -∞ to ∞). After this transformation we should be able to apply Fubini's Theorem and perform an iterated integration. But in the exponential function exp(-i\langle Cy, ζ \rangle) is still left to be dealt with and it doesn't come any where close that it could be of the form exp(-\frac{1}{2λ} \langle A^{-1}ζ, ζ \rangle) after integration as the answer suggests.. right now I'm trully stuck here.. Thanks for any help in advance!
 
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det A here means |det A| exp(i \sum_0^n{arg w_i}) where w is the eigenvalue of A.
This question even takes 20 minutes to type.. or I really suck at latex..
 
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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