sexiboi150nc

2022-09-05

Evaluate the following integral using Laplace transform
${\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-ax}\mathrm{sin}bx}{x}\phantom{\rule{thinmathspace}{0ex}}dx$

Jordyn Valdez

Your partial result does not hold. According to Laplace transform properties
${\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-ax}\mathrm{sin}bx}{x}\phantom{\rule{thinmathspace}{0ex}}dx=\mathcal{L}\left(\frac{\mathrm{sin}\mathcal{b}\mathcal{x}}{\mathcal{x}}\right)\mathcal{\left(}\mathcal{a}\mathcal{\right)}\mathcal{=}{\int }_{\mathcal{a}}^{\mathcal{+}\mathrm{\infty }}\mathcal{L}\left(\mathrm{sin}\mathcal{b}\mathcal{x}\right)\mathcal{\left(}\mathcal{p}\mathcal{\right)}\mathcal{d}\mathcal{p}\mathcal{=}{\int }_{\mathcal{a}}^{\mathcal{+}\mathrm{\infty }}\frac{\mathcal{b}}{{\mathcal{p}}^{\mathcal{2}}\mathcal{+}{\mathcal{b}}^{\mathcal{2}}}\phantom{\rule{thinmathspace}{0ex}}\mathcal{d}\mathcal{p}\mathcal{.}$

firmezas1

the standard form of LT is
$\mathcal{L}f\left(t\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$
So for the given integral you just have to calculate LT of $\frac{\mathrm{sin}bx}{x}$ then put s=a these two properties of LT may be useful
$\mathcal{L}\mathrm{sin}\left(bx\right)=\frac{b}{{b}^{2}+{s}^{2}}$
and
$\mathcal{L}\left\{\frac{f\left(t\right)}{t}\right\}={\int }_{s}^{\mathrm{\infty }}F\left(s\right)\phantom{\rule{thinmathspace}{0ex}}ds$

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