phumzaRdY

Answered

2022-11-25

How do i find the lapalace transorm of this intergral using the convolution theorem? ${\int }_{0}^{t}{e}^{-x}\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}dx$

Answer & Explanation

rezvanifanFdT

Expert

2022-11-26Added 9 answers

$f\left(t\right)={\int }_{0}^{t}{e}^{-\tau }\mathrm{cos}\tau d\tau ={e}^{-t}{\int }_{0}^{t}{e}^{t-\tau }\mathrm{cos}\tau d\tau =\left({e}^{-t}\right)\left(\left(x↦u\left(x\right){e}^{x}\right)\ast \left(x↦u\left(x\right)\mathrm{cos}x\right)\right)\left(t\right)$
$\left(\mathcal{L}f\right)\left(s\right)=\left(\mathcal{L}\left(x↦u\left(x\right){e}^{x}\right)\right)\left(s+1\right)\left(\mathcal{L}\left(x↦u\left(x\right)\mathrm{cos}x\right)\right)\left(s+1\right)$. Note the s+1 parameter to deal with the multiplication by ${e}^{-t}$

merodavandOU

Expert

2022-11-27Added 1 answers

So the transform becomes $\frac{s}{{s}^{2}+1}\frac{1}{s-1}$ with the shift $s+1$

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