Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 11

Differentiation and factorization give $f^{\prime }(x)=e^{-x^2}(1+x^2-2x^4)=e^{-x^2}(1-x^2)(1+2x^2)$. The exponential and $1+2x^2$ are positive, so the only stationary points are at $x=\pm 1$. The derivative is positive for $|x|<1$ and negative for $|x|>1$, givingThe function is odd, crosses the origin with slope one, and tends to zero from below as $x\to -\infty$ and from above as $x\to \infty$. These facts determine the sketch.

Since $f$ is odd and absolutely integrable, the two signed half-line integrals cancel:Absolute integrability follows from Gaussian decay dominating the polynomial factor; no principal-value interpretation is needed.

Put $u=x^2$, so $xdx=du/2$. Then integration by substitution givesThe last two integrals are elementary exponential moments, or values of the Gamma function.

The chain rule gives $f{f}^{\prime }=\frac{1}{2}(f^2{)}^{\prime }$. Both $f(0)$ and ${\lim }_{x\to \infty }f(x)$ vanish, so the fundamental theorem of calculus givesThe positive contribution while $f$ rises is cancelled by the negative contribution while it falls.