Past exam of the options I course of the University of Oxford / 2006 / D (Integration)

The Gaussian function is continuous and hence Lebesgue measurable. For $|x|\ge 2$, $x^2/2\ge |x|$, and thereforeThe right side has finite Lebesgue integral: its first term has integral 4 and the second is Lebesgue integrable by hypothesis. The comparison principle for the Lebesgue integral says that a measurable nonnegative function dominated by an integrable function has finite integral. Consequently the Gaussian function is Lebesgue integrable.

A real measurable function is Lebesgue integrable precisely when the integral of its absolute value is finite. For every integer $k\ge 1$, on $I_k=[k+1/6,k+5/6]$ one has $|\sin \pi x|\ge 1/2$ and $x\le k+1$. ConsequentlyThe intervals have disjoint interiors. Countable additivity of the nonnegative Lebesgue integral, or the monotone convergence theorem applied to their finite unions, givesThe harmonic series diverges, so the function is not Lebesgue integrable. Cancellation between adjacent oscillations cannot give absolute integrability.

The annulus is open and hence Lebesgue measurable; multiplying its indicator by the continuous Gaussian gives a measurable function. It is bounded by the indicator of $[-b,b{]}^2$, whose Lebesgue integral is $4b^2$. Thus it is Lebesgue integrable by comparison.

Use the change of variables formula with polar coordinates $x=r\cos \theta$, $y=r\sin \theta$. Removing one radial cut makes this a smooth one-to-one coordinate map; the cut is a null set and does not affect the integral. The absolute Jacobian determinant is $r$. By Fubini theorem,The antiderivative is $-e^{-r^2/2}$, which also makes the sign of the answer transparent.