Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 5

For $f=x\ln x$, the derivatives are $f^{\prime }=1+\ln x$ and $f^{\prime \prime }=1/x>0$. Its unique stationary point is the global minimum $(e^{-1},-e^{-1})$. For $g=\ln x/x$, $g^{\prime }=(1-\ln x)/x^2$ changes from positive to negative at $x=e$, giving the global maximum $(e,e^{-1})$. Their endpoint limits areAn intersection requires $(x^2-1)\ln x=0$, and on $x>0$ both factors vanish only at $x=1$. There $f=g=0$ and $f^{\prime }=g^{\prime }=1$, proving a common tangent line $y=x-1$. In fact,because the two numerator factors always have the same sign. Thus the curves touch at $(1,0)$ without exchanging their order.

Use $u=1/x$, so $dx=-du/u^2$ and $\ln x=-\ln u$. Reversing the integration bounds gives, without evaluating either side,For $J={\int }_0^{\infty }\ln x/(1+x^2)dx$, the same change of variables givesso $\boxed{J=0}$. These are legitimate transformations of improper integrals: $|\ln x|$ is integrable near zero, and $(\ln x)/x^2$ is integrable at infinity. In particular $J$ is absolutely convergent, so no cancellation of divergent endpoints is being assumed.