On either open half-line, part (i) says $\widehat u=-v'$. Differentiating the formula in part (iii) makes all elementary terms cancel except the <logarithm> and the bracket. Therefore
$$
\widehat u(\lambda)=-\log|\lambda|+f_+(\lambda),
\qquad \lambda>0,
$$
where
$$
f_+(\lambda)=
\int_\lambda^1\frac{e^{-ix}-1}{x}\,dx
+\int_1^\infty\frac{e^{-ix}}x\,dx,
$$
and
$$
\widehat u(\lambda)=-\log|\lambda|+f_-(\lambda),
\qquad \lambda<0,
$$
where
$$
f_-(\lambda)=
\int_{|\lambda|}^1\frac{e^{ix}-1}{x}\,dx
+\int_1^\infty\frac{e^{ix}}x\,dx.
$$
Both functions are <smooth function>[smooth] on their respective half-lines, proving the required assertion.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 327 / 1 / b / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-327/1/b/iv/solution