oliadas73
2022-09-05
Answered

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

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Yuliana Griffith

Answered 2022-09-06
Author has **6** answers

Write

$$\frac{1}{t}={\int}_{0}^{\mathrm{\infty}}dp\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}$$

Then the integral is

$$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{cos}at-\mathrm{cos}bt}{t}& ={\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}(\mathrm{cos}at-\mathrm{cos}bt){\int}_{0}^{\mathrm{\infty}}dp\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}\\ & ={\int}_{0}^{\mathrm{\infty}}dp\phantom{\rule{thinmathspace}{0ex}}({\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}at\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}-{\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}bt\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt})\\ & ={\int}_{0}^{\mathrm{\infty}}dp\phantom{\rule{thinmathspace}{0ex}}(\frac{p}{{p}^{2}+{a}^{2}}-\frac{p}{{p}^{2}+{b}^{2}})\end{array}$$

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.

$$\frac{1}{t}={\int}_{0}^{\mathrm{\infty}}dp\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}$$

Then the integral is

$$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{cos}at-\mathrm{cos}bt}{t}& ={\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}(\mathrm{cos}at-\mathrm{cos}bt){\int}_{0}^{\mathrm{\infty}}dp\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}\\ & ={\int}_{0}^{\mathrm{\infty}}dp\phantom{\rule{thinmathspace}{0ex}}({\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}at\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt}-{\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}bt\phantom{\rule{thinmathspace}{0ex}}{e}^{-pt})\\ & ={\int}_{0}^{\mathrm{\infty}}dp\phantom{\rule{thinmathspace}{0ex}}(\frac{p}{{p}^{2}+{a}^{2}}-\frac{p}{{p}^{2}+{b}^{2}})\end{array}$$

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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