Mixing
Laplace transform of integral
$begingroup$
Find the Laplace Transform of:
$int_{0}^{t} frac{Y(u)}{sqrt{t-u}}du$
I understand that $mathcal{L}{int_{0}^{t} Y(u) du} = frac{y(s)}{s}$, but I don't understand how this works when other variables are involved (in this case how do we handle the $frac{1}{sqrt{t-u}}$ term?
laplace-transform
asked Jan 27 at 2:22
Dr.DoofusDr.Doofus
12612
$endgroup$
add a comment |
$begingroup$
Find the Laplace Transform of:
$int_{0}^{t} frac{Y(u)}{sqrt{t-u}}du$
I understand that $mathcal{L}{int_{0}^{t} Y(u) du} = frac{y(s)}{s}$, but I don't understand how this works when other variables are involved (in this case how do we handle the $frac{1}{sqrt{t-u}}$ term?
laplace-transform
asked Jan 27 at 2:22
Dr.DoofusDr.Doofus
12612
$endgroup$
add a comment |
$begingroup$
Find the Laplace Transform of:
$int_{0}^{t} frac{Y(u)}{sqrt{t-u}}du$
I understand that $mathcal{L}{int_{0}^{t} Y(u) du} = frac{y(s)}{s}$, but I don't understand how this works when other variables are involved (in this case how do we handle the $frac{1}{sqrt{t-u}}$ term?
laplace-transform
asked Jan 27 at 2:22
Dr.DoofusDr.Doofus
12612
$endgroup$
Find the Laplace Transform of:
$int_{0}^{t} frac{Y(u)}{sqrt{t-u}}du$
I understand that $mathcal{L}{int_{0}^{t} Y(u) du} = frac{y(s)}{s}$, but I don't understand how this works when other variables are involved (in this case how do we handle the $frac{1}{sqrt{t-u}}$ term?
laplace-transform
laplace-transform
asked Jan 27 at 2:22
Dr.DoofusDr.Doofus
12612
asked Jan 27 at 2:22
Dr.DoofusDr.Doofus
12612
asked Jan 27 at 2:22
Dr.DoofusDr.Doofus
12612
asked Jan 27 at 2:22
Dr.DoofusDr.Doofus
12612
asked Jan 27 at 2:22
Dr.DoofusDr.Doofus
12612
12612
add a comment |
add a comment |
1 Answer
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$begingroup$
You are working with semi-primitives, nice! Well, the integral
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du $$
is the convolution between $Y(u)$ and $frac{1}{sqrt{u}}$, so its Laplace transform is simply the product between $(mathcal{L}Y)(s)$ and
$$ mathcal{L}left(frac{1}{sqrt{u}}right)(s)=frac{sqrt{pi}}{sqrt{s}}. $$
This is related to fractional calculus since a possible (but not very common) definition of a semi-primitive is
$$ (D^{-1/2} f)(x) = mathcal{L}^{-1}left[frac{1}{sqrt{s}}cdot(mathcal{L} f)(s)right](x).$$
In these terms
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du = sqrt{pi},(D^{-1/2} Y)(t).$$
It is interesting to check what happens by taking $Y$ as a Legendre or Chebyshev polynomial/function, but I do not want to spoil too much of my future work.
answered Jan 27 at 2:40
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
$begingroup$
Of course, how could I have forgotten about convolutions! Thanks, mate.
$endgroup$
– Dr.Doofus
Jan 27 at 2:48
1
$begingroup$
@Dr.Doofus: you're very welcome.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You are working with semi-primitives, nice! Well, the integral
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du $$
is the convolution between $Y(u)$ and $frac{1}{sqrt{u}}$, so its Laplace transform is simply the product between $(mathcal{L}Y)(s)$ and
$$ mathcal{L}left(frac{1}{sqrt{u}}right)(s)=frac{sqrt{pi}}{sqrt{s}}. $$
This is related to fractional calculus since a possible (but not very common) definition of a semi-primitive is
$$ (D^{-1/2} f)(x) = mathcal{L}^{-1}left[frac{1}{sqrt{s}}cdot(mathcal{L} f)(s)right](x).$$
In these terms
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du = sqrt{pi},(D^{-1/2} Y)(t).$$
It is interesting to check what happens by taking $Y$ as a Legendre or Chebyshev polynomial/function, but I do not want to spoil too much of my future work.
answered Jan 27 at 2:40
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
$begingroup$
Of course, how could I have forgotten about convolutions! Thanks, mate.
$endgroup$
– Dr.Doofus
Jan 27 at 2:48
1
$begingroup$
@Dr.Doofus: you're very welcome.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
add a comment |
$begingroup$
You are working with semi-primitives, nice! Well, the integral
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du $$
is the convolution between $Y(u)$ and $frac{1}{sqrt{u}}$, so its Laplace transform is simply the product between $(mathcal{L}Y)(s)$ and
$$ mathcal{L}left(frac{1}{sqrt{u}}right)(s)=frac{sqrt{pi}}{sqrt{s}}. $$
This is related to fractional calculus since a possible (but not very common) definition of a semi-primitive is
$$ (D^{-1/2} f)(x) = mathcal{L}^{-1}left[frac{1}{sqrt{s}}cdot(mathcal{L} f)(s)right](x).$$
In these terms
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du = sqrt{pi},(D^{-1/2} Y)(t).$$
It is interesting to check what happens by taking $Y$ as a Legendre or Chebyshev polynomial/function, but I do not want to spoil too much of my future work.
answered Jan 27 at 2:40
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
$begingroup$
Of course, how could I have forgotten about convolutions! Thanks, mate.
$endgroup$
– Dr.Doofus
Jan 27 at 2:48
1
$begingroup$
@Dr.Doofus: you're very welcome.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
add a comment |
$begingroup$
You are working with semi-primitives, nice! Well, the integral
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du $$
is the convolution between $Y(u)$ and $frac{1}{sqrt{u}}$, so its Laplace transform is simply the product between $(mathcal{L}Y)(s)$ and
$$ mathcal{L}left(frac{1}{sqrt{u}}right)(s)=frac{sqrt{pi}}{sqrt{s}}. $$
This is related to fractional calculus since a possible (but not very common) definition of a semi-primitive is
$$ (D^{-1/2} f)(x) = mathcal{L}^{-1}left[frac{1}{sqrt{s}}cdot(mathcal{L} f)(s)right](x).$$
In these terms
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du = sqrt{pi},(D^{-1/2} Y)(t).$$
It is interesting to check what happens by taking $Y$ as a Legendre or Chebyshev polynomial/function, but I do not want to spoil too much of my future work.
answered Jan 27 at 2:40
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
You are working with semi-primitives, nice! Well, the integral
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du $$
is the convolution between $Y(u)$ and $frac{1}{sqrt{u}}$, so its Laplace transform is simply the product between $(mathcal{L}Y)(s)$ and
$$ mathcal{L}left(frac{1}{sqrt{u}}right)(s)=frac{sqrt{pi}}{sqrt{s}}. $$
This is related to fractional calculus since a possible (but not very common) definition of a semi-primitive is
$$ (D^{-1/2} f)(x) = mathcal{L}^{-1}left[frac{1}{sqrt{s}}cdot(mathcal{L} f)(s)right](x).$$
In these terms
$$ int_{0}^{t}frac{Y(u)}{sqrt{t-u}},du = sqrt{pi},(D^{-1/2} Y)(t).$$
It is interesting to check what happens by taking $Y$ as a Legendre or Chebyshev polynomial/function, but I do not want to spoil too much of my future work.
answered Jan 27 at 2:40
Jack D'AurizioJack D'Aurizio
291k33284669
answered Jan 27 at 2:40
Jack D'AurizioJack D'Aurizio
291k33284669
answered Jan 27 at 2:40
Jack D'AurizioJack D'Aurizio
291k33284669
answered Jan 27 at 2:40
Jack D'AurizioJack D'Aurizio
291k33284669
291k33284669
$begingroup$
Of course, how could I have forgotten about convolutions! Thanks, mate.
$endgroup$
– Dr.Doofus
Jan 27 at 2:48
1
$begingroup$
@Dr.Doofus: you're very welcome.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
add a comment |
$begingroup$
Of course, how could I have forgotten about convolutions! Thanks, mate.
$endgroup$
– Dr.Doofus
Jan 27 at 2:48
1
$begingroup$
@Dr.Doofus: you're very welcome.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
$begingroup$
Of course, how could I have forgotten about convolutions! Thanks, mate.
$endgroup$
– Dr.Doofus
Jan 27 at 2:48
$begingroup$
Of course, how could I have forgotten about convolutions! Thanks, mate.
$endgroup$
– Dr.Doofus
Jan 27 at 2:48
1
1
$begingroup$
@Dr.Doofus: you're very welcome.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
$begingroup$
@Dr.Doofus: you're very welcome.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
add a comment |
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