Laplace transform of $frac{sin^2(t)}{t}$












1












$begingroup$


Because : $$mathscr Lleft(frac{f(t)}{t}right) = int_s^∞ F(s) ds$$



and $$mathscr Lleft(sin^2(t)right) = frac{2}{sleft(s^2 + 4right)}$$



then $$∫ frac{2}{sleft(s^2+4right)} ds = frac{1}{2}ln|s|-frac{1}{4}ln|s^2 + 4|$$



What should I do next...?



the answer that textbook gives me is $$frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)$$






laplace-transform







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  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 7:45
















1












$begingroup$


Because : $$mathscr Lleft(frac{f(t)}{t}right) = int_s^∞ F(s) ds$$



and $$mathscr Lleft(sin^2(t)right) = frac{2}{sleft(s^2 + 4right)}$$



then $$∫ frac{2}{sleft(s^2+4right)} ds = frac{1}{2}ln|s|-frac{1}{4}ln|s^2 + 4|$$



What should I do next...?



the answer that textbook gives me is $$frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)$$






laplace-transform







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 7:45














1












1








1





$begingroup$


Because : $$mathscr Lleft(frac{f(t)}{t}right) = int_s^∞ F(s) ds$$



and $$mathscr Lleft(sin^2(t)right) = frac{2}{sleft(s^2 + 4right)}$$



then $$∫ frac{2}{sleft(s^2+4right)} ds = frac{1}{2}ln|s|-frac{1}{4}ln|s^2 + 4|$$



What should I do next...?



the answer that textbook gives me is $$frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)$$






laplace-transform







share|cite|improve this question











$endgroup$




Because : $$mathscr Lleft(frac{f(t)}{t}right) = int_s^∞ F(s) ds$$



and $$mathscr Lleft(sin^2(t)right) = frac{2}{sleft(s^2 + 4right)}$$



then $$∫ frac{2}{sleft(s^2+4right)} ds = frac{1}{2}ln|s|-frac{1}{4}ln|s^2 + 4|$$



What should I do next...?



the answer that textbook gives me is $$frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)$$






laplace-transform




laplace-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 8:50









DavidG

2,5061726




2,5061726










asked Dec 23 '18 at 7:43









FOMTFOMT

133




133








  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 7:45














  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 7:45








2




2




$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 7:45




$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 7:45










1 Answer
1






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oldest

votes


















1












$begingroup$

Using the double angle identity:



begin{equation}
sin^2(t) = frac{1 - cos(2t)}{2}
end{equation}



Thus,



begin{equation}
mathscr{L}left[sin^2(t)right] = frac{1}{2s} - frac{s}{2left(s^2 + 4right)}
end{equation}



We now employ the property of Laplace Transforms:



begin{equation}
mathscr{L}left[ frac{f(t)}{t} right] = int_{s}^{infty}mathscr{L}left[ f(t) right]_u:ds
end{equation}



Thus,



begin{align}
mathscr{L}left[ frac{sin^2(t)}{t} right] &= int_{s}^{infty}left[ frac{1}{2u} - frac{u}{2left(u^2 + 4right)} right]:du = left[ frac{1}{2}ln(u) - frac{1}{4}lnleft(u^2 + 4right)right]_{s}^{infty}\
& = left[frac{1}{4}lnleft(frac{u^2}{u^2 + 4} right) right]_{s}^{infty} = frac{1}{4}left[0 - lnleft(frac{s^2}{s^2 + 4} right) right] = - frac{1}{4}lnleft(frac{s^2}{s^2 + 4}right)\
& = frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)
end{align}






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  • $begingroup$
    If you're looking for other methods, I recommend taking away the 'Answered' status on my comment (many ignore answered questions). Up to you of course! Merry xmas!
    $endgroup$
    – DavidG
    Dec 23 '18 at 10:17











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

Using the double angle identity:



begin{equation}
sin^2(t) = frac{1 - cos(2t)}{2}
end{equation}



Thus,



begin{equation}
mathscr{L}left[sin^2(t)right] = frac{1}{2s} - frac{s}{2left(s^2 + 4right)}
end{equation}



We now employ the property of Laplace Transforms:



begin{equation}
mathscr{L}left[ frac{f(t)}{t} right] = int_{s}^{infty}mathscr{L}left[ f(t) right]_u:ds
end{equation}



Thus,



begin{align}
mathscr{L}left[ frac{sin^2(t)}{t} right] &= int_{s}^{infty}left[ frac{1}{2u} - frac{u}{2left(u^2 + 4right)} right]:du = left[ frac{1}{2}ln(u) - frac{1}{4}lnleft(u^2 + 4right)right]_{s}^{infty}\
& = left[frac{1}{4}lnleft(frac{u^2}{u^2 + 4} right) right]_{s}^{infty} = frac{1}{4}left[0 - lnleft(frac{s^2}{s^2 + 4} right) right] = - frac{1}{4}lnleft(frac{s^2}{s^2 + 4}right)\
& = frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If you're looking for other methods, I recommend taking away the 'Answered' status on my comment (many ignore answered questions). Up to you of course! Merry xmas!
    $endgroup$
    – DavidG
    Dec 23 '18 at 10:17
















1












$begingroup$

Using the double angle identity:



begin{equation}
sin^2(t) = frac{1 - cos(2t)}{2}
end{equation}



Thus,



begin{equation}
mathscr{L}left[sin^2(t)right] = frac{1}{2s} - frac{s}{2left(s^2 + 4right)}
end{equation}



We now employ the property of Laplace Transforms:



begin{equation}
mathscr{L}left[ frac{f(t)}{t} right] = int_{s}^{infty}mathscr{L}left[ f(t) right]_u:ds
end{equation}



Thus,



begin{align}
mathscr{L}left[ frac{sin^2(t)}{t} right] &= int_{s}^{infty}left[ frac{1}{2u} - frac{u}{2left(u^2 + 4right)} right]:du = left[ frac{1}{2}ln(u) - frac{1}{4}lnleft(u^2 + 4right)right]_{s}^{infty}\
& = left[frac{1}{4}lnleft(frac{u^2}{u^2 + 4} right) right]_{s}^{infty} = frac{1}{4}left[0 - lnleft(frac{s^2}{s^2 + 4} right) right] = - frac{1}{4}lnleft(frac{s^2}{s^2 + 4}right)\
& = frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If you're looking for other methods, I recommend taking away the 'Answered' status on my comment (many ignore answered questions). Up to you of course! Merry xmas!
    $endgroup$
    – DavidG
    Dec 23 '18 at 10:17














1












1








1





$begingroup$

Using the double angle identity:



begin{equation}
sin^2(t) = frac{1 - cos(2t)}{2}
end{equation}



Thus,



begin{equation}
mathscr{L}left[sin^2(t)right] = frac{1}{2s} - frac{s}{2left(s^2 + 4right)}
end{equation}



We now employ the property of Laplace Transforms:



begin{equation}
mathscr{L}left[ frac{f(t)}{t} right] = int_{s}^{infty}mathscr{L}left[ f(t) right]_u:ds
end{equation}



Thus,



begin{align}
mathscr{L}left[ frac{sin^2(t)}{t} right] &= int_{s}^{infty}left[ frac{1}{2u} - frac{u}{2left(u^2 + 4right)} right]:du = left[ frac{1}{2}ln(u) - frac{1}{4}lnleft(u^2 + 4right)right]_{s}^{infty}\
& = left[frac{1}{4}lnleft(frac{u^2}{u^2 + 4} right) right]_{s}^{infty} = frac{1}{4}left[0 - lnleft(frac{s^2}{s^2 + 4} right) right] = - frac{1}{4}lnleft(frac{s^2}{s^2 + 4}right)\
& = frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)
end{align}






share|cite|improve this answer









$endgroup$



Using the double angle identity:



begin{equation}
sin^2(t) = frac{1 - cos(2t)}{2}
end{equation}



Thus,



begin{equation}
mathscr{L}left[sin^2(t)right] = frac{1}{2s} - frac{s}{2left(s^2 + 4right)}
end{equation}



We now employ the property of Laplace Transforms:



begin{equation}
mathscr{L}left[ frac{f(t)}{t} right] = int_{s}^{infty}mathscr{L}left[ f(t) right]_u:ds
end{equation}



Thus,



begin{align}
mathscr{L}left[ frac{sin^2(t)}{t} right] &= int_{s}^{infty}left[ frac{1}{2u} - frac{u}{2left(u^2 + 4right)} right]:du = left[ frac{1}{2}ln(u) - frac{1}{4}lnleft(u^2 + 4right)right]_{s}^{infty}\
& = left[frac{1}{4}lnleft(frac{u^2}{u^2 + 4} right) right]_{s}^{infty} = frac{1}{4}left[0 - lnleft(frac{s^2}{s^2 + 4} right) right] = - frac{1}{4}lnleft(frac{s^2}{s^2 + 4}right)\
& = frac{1}{4}lnleft(frac{s^2 + 4}{s^2}right)
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 8:07









DavidGDavidG

2,5061726




2,5061726












  • $begingroup$
    If you're looking for other methods, I recommend taking away the 'Answered' status on my comment (many ignore answered questions). Up to you of course! Merry xmas!
    $endgroup$
    – DavidG
    Dec 23 '18 at 10:17


















  • $begingroup$
    If you're looking for other methods, I recommend taking away the 'Answered' status on my comment (many ignore answered questions). Up to you of course! Merry xmas!
    $endgroup$
    – DavidG
    Dec 23 '18 at 10:17
















$begingroup$
If you're looking for other methods, I recommend taking away the 'Answered' status on my comment (many ignore answered questions). Up to you of course! Merry xmas!
$endgroup$
– DavidG
Dec 23 '18 at 10:17




$begingroup$
If you're looking for other methods, I recommend taking away the 'Answered' status on my comment (many ignore answered questions). Up to you of course! Merry xmas!
$endgroup$
– DavidG
Dec 23 '18 at 10:17


















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