An integral of Wolstenholme:$\int_0^{+\infty}\frac{\sum_1^n A_k\cos{a_k x}}{x}\mathrm {d} x$ where $\sum A_k=0$ and $a_k>0$

Question

The book by Whittaker and Watson says it's equal to $-\sum_{k=1}^n A_k \log {a_k}$, and attibutes it to Wolstenholme.

I believe this readily reduces to the simpler case of evaluating $\displaystyle \int_0^{+\infty} \frac{\cos ax -\cos bx}{x}\mathrm{d}x$.

My attempt (in vain) : complexify the cosine function, since $\displaystyle \int_0^{+\infty} \frac{\cos ax -\cos bx}{x}\mathrm{d}x=\displaystyle \int_0^{+\infty} \frac{e^{ axi} -e^{bxi}}{x}\mathrm{d}x$. But how does one evaluate the latter?

Answer 1

This falls into a more general class of integrals called Frullani's integral. It takes the general form

$$ \int_{0}^{\infty} \frac{f(ax) - f(bx)}{x} \, dx = C_{f} (\log a - \log b) $$

for some constant $C_{f}$, and it has been evaluated for a fairly large class of functions. In this case, a simple trick works: for $h > 0$,

\begin{align*} \int_{h}^{\infty} \frac{\cos ax - \cos bx}{x} \, dx &= \int_{h}^{\infty} \frac{\cos ax}{x} \, dx - \int_{h}^{\infty} \frac{\cos bx}{x} \, dx \\ &= \int_{ah}^{\infty} \frac{\cos x}{x} \, dx - \int_{bh}^{\infty} \frac{\cos x}{x} \, dx \\ &= \int_{ah}^{bh} \frac{\cos x}{x} \, dx \\ &= \int_{a}^{b} \frac{\cos (hx)}{x} \, dx \end{align*}

Thus taking $h \to 0^{+}$, it follows that

$$ \int_{0}^{\infty} \frac{\cos ax - \cos bx}{x} \, dx = -(\log a - \log b). $$