Inverse Laplace transform of $1/(1+sqrt{1+s^2})$
A seemingly simple function,
$F(s)=frac{1}{1+sqrt{1+s^2}}$.
But what is its inverse Laplace transform?
If there is not a closed form, can I get some long/short time asymptotic?
It must have something to do with the Bessel J function.
laplace-transform
asked Nov 29 '18 at 20:16
user621127user621127
61
add a comment |
A seemingly simple function,
$F(s)=frac{1}{1+sqrt{1+s^2}}$.
But what is its inverse Laplace transform?
If there is not a closed form, can I get some long/short time asymptotic?
It must have something to do with the Bessel J function.
laplace-transform
asked Nov 29 '18 at 20:16
user621127user621127
61
add a comment |
A seemingly simple function,
$F(s)=frac{1}{1+sqrt{1+s^2}}$.
But what is its inverse Laplace transform?
If there is not a closed form, can I get some long/short time asymptotic?
It must have something to do with the Bessel J function.
laplace-transform
asked Nov 29 '18 at 20:16
user621127user621127
61
A seemingly simple function,
$F(s)=frac{1}{1+sqrt{1+s^2}}$.
But what is its inverse Laplace transform?
If there is not a closed form, can I get some long/short time asymptotic?
It must have something to do with the Bessel J function.
laplace-transform
laplace-transform
asked Nov 29 '18 at 20:16
user621127user621127
61
asked Nov 29 '18 at 20:16
user621127user621127
61
asked Nov 29 '18 at 20:16
user621127user621127
61
asked Nov 29 '18 at 20:16
user621127user621127
61
asked Nov 29 '18 at 20:16
user621127user621127
61
61
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
We can use the trick of writing
$$frac 1 {1 + sqrt {1 + s^2}} =
frac 1 s sqrt {1 + frac 1 {s^2}} - frac 1 {s^2},$$
which holds for $s$ in the right half-plane. Then
$$mathcal L^{-1} !{left[ frac 1 {1 + sqrt {1 + s^2}} right]} =
-mathcal L^{-1}[s^{-2}] +
sum_{k geq 0} binom {1/2} k mathcal L^{-1}[s^{-2 k - 1}] = \
-t + sum_{k geq 0} binom {1/2} k frac {t^{2 k}} {Gamma(2 k + 1)} =
{_1hspace{-2px}F_2} {left( -frac 1 2; frac 1 2, 1; -frac {t^2} 4 right)} - t.$$
The hypergeometric function can be converted to a combination of Bessel and Struve functions.
answered Dec 1 '18 at 4:13
MaximMaxim
4,6081219
Thank you so much, this is brilliant!
– user621127
Dec 2 '18 at 7:51
May I ask a follow up though. So what happens if my function becomes $$frac{1}{c+sqrt{1+s^2}}$$, where $c$ is a constant other than 1. Then the "partial fraction" still works, the binomial theorem still works, but the product is not as nice and we don't just get 1/s that can be absorbed. In this case, is there a way other than convolution?
– user621127
Dec 2 '18 at 7:54
For $c neq pm 1$, the inverse transform will contain an integral of $e^{a t} J_1(t)/t$ (which is the convolution that you mentioned). Most likely some very general special functions will be required to express it in closed form.
– Maxim
Dec 2 '18 at 11:04
add a comment |
Hint
Let the inverse Laplace be $f(t)$ . Then what is the Laplace of $frac{f(t)}{t}$
answered Dec 1 '18 at 4:01
user471651user471651
513215
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We can use the trick of writing
$$frac 1 {1 + sqrt {1 + s^2}} =
frac 1 s sqrt {1 + frac 1 {s^2}} - frac 1 {s^2},$$
which holds for $s$ in the right half-plane. Then
$$mathcal L^{-1} !{left[ frac 1 {1 + sqrt {1 + s^2}} right]} =
-mathcal L^{-1}[s^{-2}] +
sum_{k geq 0} binom {1/2} k mathcal L^{-1}[s^{-2 k - 1}] = \
-t + sum_{k geq 0} binom {1/2} k frac {t^{2 k}} {Gamma(2 k + 1)} =
{_1hspace{-2px}F_2} {left( -frac 1 2; frac 1 2, 1; -frac {t^2} 4 right)} - t.$$
The hypergeometric function can be converted to a combination of Bessel and Struve functions.
answered Dec 1 '18 at 4:13
MaximMaxim
4,6081219
Thank you so much, this is brilliant!
– user621127
Dec 2 '18 at 7:51
May I ask a follow up though. So what happens if my function becomes $$frac{1}{c+sqrt{1+s^2}}$$, where $c$ is a constant other than 1. Then the "partial fraction" still works, the binomial theorem still works, but the product is not as nice and we don't just get 1/s that can be absorbed. In this case, is there a way other than convolution?
– user621127
Dec 2 '18 at 7:54
For $c neq pm 1$, the inverse transform will contain an integral of $e^{a t} J_1(t)/t$ (which is the convolution that you mentioned). Most likely some very general special functions will be required to express it in closed form.
– Maxim
Dec 2 '18 at 11:04
add a comment |
We can use the trick of writing
$$frac 1 {1 + sqrt {1 + s^2}} =
frac 1 s sqrt {1 + frac 1 {s^2}} - frac 1 {s^2},$$
which holds for $s$ in the right half-plane. Then
$$mathcal L^{-1} !{left[ frac 1 {1 + sqrt {1 + s^2}} right]} =
-mathcal L^{-1}[s^{-2}] +
sum_{k geq 0} binom {1/2} k mathcal L^{-1}[s^{-2 k - 1}] = \
-t + sum_{k geq 0} binom {1/2} k frac {t^{2 k}} {Gamma(2 k + 1)} =
{_1hspace{-2px}F_2} {left( -frac 1 2; frac 1 2, 1; -frac {t^2} 4 right)} - t.$$
The hypergeometric function can be converted to a combination of Bessel and Struve functions.
answered Dec 1 '18 at 4:13
MaximMaxim
4,6081219
Thank you so much, this is brilliant!
– user621127
Dec 2 '18 at 7:51
May I ask a follow up though. So what happens if my function becomes $$frac{1}{c+sqrt{1+s^2}}$$, where $c$ is a constant other than 1. Then the "partial fraction" still works, the binomial theorem still works, but the product is not as nice and we don't just get 1/s that can be absorbed. In this case, is there a way other than convolution?
– user621127
Dec 2 '18 at 7:54
For $c neq pm 1$, the inverse transform will contain an integral of $e^{a t} J_1(t)/t$ (which is the convolution that you mentioned). Most likely some very general special functions will be required to express it in closed form.
– Maxim
Dec 2 '18 at 11:04
add a comment |
We can use the trick of writing
$$frac 1 {1 + sqrt {1 + s^2}} =
frac 1 s sqrt {1 + frac 1 {s^2}} - frac 1 {s^2},$$
which holds for $s$ in the right half-plane. Then
$$mathcal L^{-1} !{left[ frac 1 {1 + sqrt {1 + s^2}} right]} =
-mathcal L^{-1}[s^{-2}] +
sum_{k geq 0} binom {1/2} k mathcal L^{-1}[s^{-2 k - 1}] = \
-t + sum_{k geq 0} binom {1/2} k frac {t^{2 k}} {Gamma(2 k + 1)} =
{_1hspace{-2px}F_2} {left( -frac 1 2; frac 1 2, 1; -frac {t^2} 4 right)} - t.$$
The hypergeometric function can be converted to a combination of Bessel and Struve functions.
answered Dec 1 '18 at 4:13
MaximMaxim
4,6081219
We can use the trick of writing
$$frac 1 {1 + sqrt {1 + s^2}} =
frac 1 s sqrt {1 + frac 1 {s^2}} - frac 1 {s^2},$$
which holds for $s$ in the right half-plane. Then
$$mathcal L^{-1} !{left[ frac 1 {1 + sqrt {1 + s^2}} right]} =
-mathcal L^{-1}[s^{-2}] +
sum_{k geq 0} binom {1/2} k mathcal L^{-1}[s^{-2 k - 1}] = \
-t + sum_{k geq 0} binom {1/2} k frac {t^{2 k}} {Gamma(2 k + 1)} =
{_1hspace{-2px}F_2} {left( -frac 1 2; frac 1 2, 1; -frac {t^2} 4 right)} - t.$$
The hypergeometric function can be converted to a combination of Bessel and Struve functions.
answered Dec 1 '18 at 4:13
MaximMaxim
4,6081219
answered Dec 1 '18 at 4:13
MaximMaxim
4,6081219
answered Dec 1 '18 at 4:13
MaximMaxim
4,6081219
answered Dec 1 '18 at 4:13
MaximMaxim
4,6081219
4,6081219
Thank you so much, this is brilliant!
– user621127
Dec 2 '18 at 7:51
May I ask a follow up though. So what happens if my function becomes $$frac{1}{c+sqrt{1+s^2}}$$, where $c$ is a constant other than 1. Then the "partial fraction" still works, the binomial theorem still works, but the product is not as nice and we don't just get 1/s that can be absorbed. In this case, is there a way other than convolution?
– user621127
Dec 2 '18 at 7:54
For $c neq pm 1$, the inverse transform will contain an integral of $e^{a t} J_1(t)/t$ (which is the convolution that you mentioned). Most likely some very general special functions will be required to express it in closed form.
– Maxim
Dec 2 '18 at 11:04
add a comment |
Thank you so much, this is brilliant!
– user621127
Dec 2 '18 at 7:51
May I ask a follow up though. So what happens if my function becomes $$frac{1}{c+sqrt{1+s^2}}$$, where $c$ is a constant other than 1. Then the "partial fraction" still works, the binomial theorem still works, but the product is not as nice and we don't just get 1/s that can be absorbed. In this case, is there a way other than convolution?
– user621127
Dec 2 '18 at 7:54
For $c neq pm 1$, the inverse transform will contain an integral of $e^{a t} J_1(t)/t$ (which is the convolution that you mentioned). Most likely some very general special functions will be required to express it in closed form.
– Maxim
Dec 2 '18 at 11:04
Thank you so much, this is brilliant!
– user621127
Dec 2 '18 at 7:51
Thank you so much, this is brilliant!
– user621127
Dec 2 '18 at 7:51
May I ask a follow up though. So what happens if my function becomes $$frac{1}{c+sqrt{1+s^2}}$$, where $c$ is a constant other than 1. Then the "partial fraction" still works, the binomial theorem still works, but the product is not as nice and we don't just get 1/s that can be absorbed. In this case, is there a way other than convolution?
– user621127
Dec 2 '18 at 7:54
May I ask a follow up though. So what happens if my function becomes $$frac{1}{c+sqrt{1+s^2}}$$, where $c$ is a constant other than 1. Then the "partial fraction" still works, the binomial theorem still works, but the product is not as nice and we don't just get 1/s that can be absorbed. In this case, is there a way other than convolution?
– user621127
Dec 2 '18 at 7:54
For $c neq pm 1$, the inverse transform will contain an integral of $e^{a t} J_1(t)/t$ (which is the convolution that you mentioned). Most likely some very general special functions will be required to express it in closed form.
– Maxim
Dec 2 '18 at 11:04
For $c neq pm 1$, the inverse transform will contain an integral of $e^{a t} J_1(t)/t$ (which is the convolution that you mentioned). Most likely some very general special functions will be required to express it in closed form.
– Maxim
Dec 2 '18 at 11:04
add a comment |
Hint
Let the inverse Laplace be $f(t)$ . Then what is the Laplace of $frac{f(t)}{t}$
answered Dec 1 '18 at 4:01
user471651user471651
513215
add a comment |
Hint
Let the inverse Laplace be $f(t)$ . Then what is the Laplace of $frac{f(t)}{t}$
answered Dec 1 '18 at 4:01
user471651user471651
513215
add a comment |
Hint
Let the inverse Laplace be $f(t)$ . Then what is the Laplace of $frac{f(t)}{t}$
answered Dec 1 '18 at 4:01
user471651user471651
513215
Hint
Let the inverse Laplace be $f(t)$ . Then what is the Laplace of $frac{f(t)}{t}$
answered Dec 1 '18 at 4:01
user471651user471651
513215
answered Dec 1 '18 at 4:01
user471651user471651
513215
answered Dec 1 '18 at 4:01
user471651user471651
513215
answered Dec 1 '18 at 4:01
user471651user471651
513215
513215
add a comment |
add a comment |
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