Find the integral of this function

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    Function Integral
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SUMMARY

The integral of the function $f(x,y) = \frac{1}{(x^2 + y^2 + 1)^{\frac{3}{2}}}$ over the closed ball $\overline{B(a, 2)}$ is evaluated as $a \rightarrow \infty$. The discussion confirms that the limit leads to the result $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) \, dy \, dx = 2\pi$. A clarification was made regarding the variable $a$, which represents the center of the disk, not its radius.

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kalvin
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Find the integral of the function $f(x,y) = \frac{1}{(x^2 + y^2 + 1)^{\frac{3}{2}}} $
over the closed ball $\overline{B(a, 2)}$(i.e disk with radius 2 centered at point a). Letting $a \rightarrow \infty$, show that:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(x,y) dydx = 2\pi$
 
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kalvin said:
Find the integral of the function $f(x,y) = \frac{1}{(x^2 + y^2 + 1)^{\frac{3}{2}}} $
over the closed ball $\overline{B(a, 2)}$(i.e disk with radius 2 centered at point a). Letting $a \rightarrow \infty$, show that:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(x,y) dydx = 2\pi$
:confused: That wording cannot be correct: $a$ must surely be the radius of the disc, not its centre?
 

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