2022-09-05

Show that $\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{t}\left(\mathrm{cos}\left(at\right)-\mathrm{cos}\left(bt\right)\right)dt=\mathrm{ln}\left(b/a\right),\phantom{\rule{thinmathspace}{0ex}}a,b>0$

Yuliana Griffith

Write
$\frac{1}{t}={\int }_{0}^{\mathrm{\infty }}dp\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}$
Then the integral is
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{cos}at-\mathrm{cos}bt}{t}& ={\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}\left(\mathrm{cos}at-\mathrm{cos}bt\right){\int }_{0}^{\mathrm{\infty }}dp\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}\\ & ={\int }_{0}^{\mathrm{\infty }}dp\phantom{\rule{thinmathspace}{0ex}}\left({\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}at\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}-{\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}bt\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}\right)\\ & ={\int }_{0}^{\mathrm{\infty }}dp\phantom{\rule{thinmathspace}{0ex}}\left(\frac{p}{{p}^{2}+{a}^{2}}-\frac{p}{{p}^{2}+{b}^{2}}\right)\end{array}$
It seems you got the rest. Note that, in the second line, we can reverse the order of integration because the integrals involved are convergent.

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