Suppose $w$ is locally integrable and
$$
w(x)=\frac{c_+}{x}+O(x^{-2})\quad(x\to+\infty),
\qquad
w(x)=\frac{c_-}{x}+O(x^{-2})\quad(x\to-\infty).
$$
Then $\widehat w(\lambda)+(c_+-c_-)\log|\lambda|$ has finite limits at zero from both sides. With the convention $\widehat w(\lambda)=\int e^{-i\lambda x}w(x)dx$, the positive-frequency limit minus the negative-frequency limit is
$$
-i\pi(c_++c_-).
$$
Subtracting suitable reflected copies of $x^{-1}\mathbf1_{(1,\infty)}(x)$ leaves an $L^1$ function, whose Fourier transform is continuous; the jump then follows from the <Dirichlet integral>.

Fourier transform of a function with reciprocal tails

For $\lambda>0$, subtracting the two formulas from part (iv) yields
$$
f_+(\lambda)-f_-(-\lambda)
=\int_\lambda^\infty
\frac{e^{-ix}-e^{ix}}x\,dx
=-2i\int_\lambda^\infty\frac{\sin x}{x}\,dx.
$$
Using the stated <Dirichlet integral> and taking the <one-sided limit> gives
$$
\boxed{
\lim_{\lambda\downarrow0}
\bigl[f_+(\lambda)-f_-(-\lambda)\bigr]
=-2i\frac\pi2=-i\pi}.
$$

Solution

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

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The Dirichlet integral is the conditionally convergent improper integral
$$
\int_0^\infty\frac{\sin x}{x}\,dx=\frac\pi2.
$$
It controls the jump behavior of Fourier transforms and Fourier inversion at discontinuities.


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