Laplace Transform of Complementary Error Function












0












$begingroup$


I need to apply one Laplace transform formula while I have no idea how to prove it:
$$int_0^infty e^{-st} e^{a k} e^{a^2 t}
operatorname{erfc} left( a sqrt{t} + frac{k}{2 sqrt{t}} right) dt =
frac{e^{-k sqrt{s}}}{sqrt{s} (sqrt{s}+a)},
quad k>0 land a in mathbb{C}, $$
where $operatorname{erfc}(t) = frac{2}{sqrt{pi}} int_t^infty e^{-x^2} dx$.



Could anyone help me with it? Thanks in advance.






laplace-transform integral-transforms error-function







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migrated from mathematica.stackexchange.com Dec 9 '18 at 19:09


This question came from our site for users of Wolfram Mathematica.














  • 1




    $begingroup$
    After dropping the constant factor $e^{a k}$ and differentiating wrt $k$, we need to show that $$int mathcal L_{t to s} {left[ -frac {e^{-a k - k^2/(4 t)}} {sqrt{pi t}} right]} dk = int -frac {e^{-a k - k sqrt s}} {sqrt s} dk = frac {e^{-a k - k sqrt s}} {sqrt s (sqrt s + a)}.$$
    $endgroup$
    – Maxim
    Dec 17 '18 at 11:48












  • $begingroup$
    Thanks for that. I like your method. My way is simply doing this integral step by step. However, an initial condition is required in your way. I take $k=0$, in that case, we need a>0 for the Laplace transform of $e^{a^2 t}erfc(asqrt{t})$. I am wondering how to extend "a" to the whole complex plane.
    $endgroup$
    – gouwangzhangdong
    Dec 19 '18 at 2:16








  • 1




    $begingroup$
    You can extend your formula to $k geq 0 land a in mathbb C$. $int_0^infty e^{a^2 t} operatorname{erfc}(a sqrt t) e^{-s t} dt$ converges for $$(operatorname{Re} a geq 0 land operatorname{Re} s > 0) lor (operatorname{Re} a < 0 land operatorname{Re} s > max(operatorname{Re}a^2, 0)),$$ in agreement with the location of the zeroes of $sqrt s (sqrt s + a)$. Another way is to multiply back by $e^{a k}$ and take $k = infty$ to show that the integration constant is zero.
    $endgroup$
    – Maxim
    Dec 19 '18 at 14:52










  • $begingroup$
    Many thanks.Fantastic. I really should not forget this analytic continuity issue. I have another question related to the complementary error function, could you please take a look? math.stackexchange.com/questions/3047021/…
    $endgroup$
    – gouwangzhangdong
    Dec 20 '18 at 0:43


















0












$begingroup$


I need to apply one Laplace transform formula while I have no idea how to prove it:
$$int_0^infty e^{-st} e^{a k} e^{a^2 t}
operatorname{erfc} left( a sqrt{t} + frac{k}{2 sqrt{t}} right) dt =
frac{e^{-k sqrt{s}}}{sqrt{s} (sqrt{s}+a)},
quad k>0 land a in mathbb{C}, $$
where $operatorname{erfc}(t) = frac{2}{sqrt{pi}} int_t^infty e^{-x^2} dx$.



Could anyone help me with it? Thanks in advance.






laplace-transform integral-transforms error-function







share|cite|improve this question











$endgroup$



migrated from mathematica.stackexchange.com Dec 9 '18 at 19:09


This question came from our site for users of Wolfram Mathematica.














  • 1




    $begingroup$
    After dropping the constant factor $e^{a k}$ and differentiating wrt $k$, we need to show that $$int mathcal L_{t to s} {left[ -frac {e^{-a k - k^2/(4 t)}} {sqrt{pi t}} right]} dk = int -frac {e^{-a k - k sqrt s}} {sqrt s} dk = frac {e^{-a k - k sqrt s}} {sqrt s (sqrt s + a)}.$$
    $endgroup$
    – Maxim
    Dec 17 '18 at 11:48












  • $begingroup$
    Thanks for that. I like your method. My way is simply doing this integral step by step. However, an initial condition is required in your way. I take $k=0$, in that case, we need a>0 for the Laplace transform of $e^{a^2 t}erfc(asqrt{t})$. I am wondering how to extend "a" to the whole complex plane.
    $endgroup$
    – gouwangzhangdong
    Dec 19 '18 at 2:16








  • 1




    $begingroup$
    You can extend your formula to $k geq 0 land a in mathbb C$. $int_0^infty e^{a^2 t} operatorname{erfc}(a sqrt t) e^{-s t} dt$ converges for $$(operatorname{Re} a geq 0 land operatorname{Re} s > 0) lor (operatorname{Re} a < 0 land operatorname{Re} s > max(operatorname{Re}a^2, 0)),$$ in agreement with the location of the zeroes of $sqrt s (sqrt s + a)$. Another way is to multiply back by $e^{a k}$ and take $k = infty$ to show that the integration constant is zero.
    $endgroup$
    – Maxim
    Dec 19 '18 at 14:52










  • $begingroup$
    Many thanks.Fantastic. I really should not forget this analytic continuity issue. I have another question related to the complementary error function, could you please take a look? math.stackexchange.com/questions/3047021/…
    $endgroup$
    – gouwangzhangdong
    Dec 20 '18 at 0:43
















0












0








0


1



$begingroup$


I need to apply one Laplace transform formula while I have no idea how to prove it:
$$int_0^infty e^{-st} e^{a k} e^{a^2 t}
operatorname{erfc} left( a sqrt{t} + frac{k}{2 sqrt{t}} right) dt =
frac{e^{-k sqrt{s}}}{sqrt{s} (sqrt{s}+a)},
quad k>0 land a in mathbb{C}, $$
where $operatorname{erfc}(t) = frac{2}{sqrt{pi}} int_t^infty e^{-x^2} dx$.



Could anyone help me with it? Thanks in advance.






laplace-transform integral-transforms error-function







share|cite|improve this question











$endgroup$




I need to apply one Laplace transform formula while I have no idea how to prove it:
$$int_0^infty e^{-st} e^{a k} e^{a^2 t}
operatorname{erfc} left( a sqrt{t} + frac{k}{2 sqrt{t}} right) dt =
frac{e^{-k sqrt{s}}}{sqrt{s} (sqrt{s}+a)},
quad k>0 land a in mathbb{C}, $$
where $operatorname{erfc}(t) = frac{2}{sqrt{pi}} int_t^infty e^{-x^2} dx$.



Could anyone help me with it? Thanks in advance.






laplace-transform integral-transforms error-function




laplace-transform integral-transforms error-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 11:54









Maxim

5,9881221




5,9881221










asked Dec 8 '18 at 11:21









gouwangzhangdonggouwangzhangdong

888




888




migrated from mathematica.stackexchange.com Dec 9 '18 at 19:09


This question came from our site for users of Wolfram Mathematica.









migrated from mathematica.stackexchange.com Dec 9 '18 at 19:09


This question came from our site for users of Wolfram Mathematica.










  • 1




    $begingroup$
    After dropping the constant factor $e^{a k}$ and differentiating wrt $k$, we need to show that $$int mathcal L_{t to s} {left[ -frac {e^{-a k - k^2/(4 t)}} {sqrt{pi t}} right]} dk = int -frac {e^{-a k - k sqrt s}} {sqrt s} dk = frac {e^{-a k - k sqrt s}} {sqrt s (sqrt s + a)}.$$
    $endgroup$
    – Maxim
    Dec 17 '18 at 11:48












  • $begingroup$
    Thanks for that. I like your method. My way is simply doing this integral step by step. However, an initial condition is required in your way. I take $k=0$, in that case, we need a>0 for the Laplace transform of $e^{a^2 t}erfc(asqrt{t})$. I am wondering how to extend "a" to the whole complex plane.
    $endgroup$
    – gouwangzhangdong
    Dec 19 '18 at 2:16








  • 1




    $begingroup$
    You can extend your formula to $k geq 0 land a in mathbb C$. $int_0^infty e^{a^2 t} operatorname{erfc}(a sqrt t) e^{-s t} dt$ converges for $$(operatorname{Re} a geq 0 land operatorname{Re} s > 0) lor (operatorname{Re} a < 0 land operatorname{Re} s > max(operatorname{Re}a^2, 0)),$$ in agreement with the location of the zeroes of $sqrt s (sqrt s + a)$. Another way is to multiply back by $e^{a k}$ and take $k = infty$ to show that the integration constant is zero.
    $endgroup$
    – Maxim
    Dec 19 '18 at 14:52










  • $begingroup$
    Many thanks.Fantastic. I really should not forget this analytic continuity issue. I have another question related to the complementary error function, could you please take a look? math.stackexchange.com/questions/3047021/…
    $endgroup$
    – gouwangzhangdong
    Dec 20 '18 at 0:43
















  • 1




    $begingroup$
    After dropping the constant factor $e^{a k}$ and differentiating wrt $k$, we need to show that $$int mathcal L_{t to s} {left[ -frac {e^{-a k - k^2/(4 t)}} {sqrt{pi t}} right]} dk = int -frac {e^{-a k - k sqrt s}} {sqrt s} dk = frac {e^{-a k - k sqrt s}} {sqrt s (sqrt s + a)}.$$
    $endgroup$
    – Maxim
    Dec 17 '18 at 11:48












  • $begingroup$
    Thanks for that. I like your method. My way is simply doing this integral step by step. However, an initial condition is required in your way. I take $k=0$, in that case, we need a>0 for the Laplace transform of $e^{a^2 t}erfc(asqrt{t})$. I am wondering how to extend "a" to the whole complex plane.
    $endgroup$
    – gouwangzhangdong
    Dec 19 '18 at 2:16








  • 1




    $begingroup$
    You can extend your formula to $k geq 0 land a in mathbb C$. $int_0^infty e^{a^2 t} operatorname{erfc}(a sqrt t) e^{-s t} dt$ converges for $$(operatorname{Re} a geq 0 land operatorname{Re} s > 0) lor (operatorname{Re} a < 0 land operatorname{Re} s > max(operatorname{Re}a^2, 0)),$$ in agreement with the location of the zeroes of $sqrt s (sqrt s + a)$. Another way is to multiply back by $e^{a k}$ and take $k = infty$ to show that the integration constant is zero.
    $endgroup$
    – Maxim
    Dec 19 '18 at 14:52










  • $begingroup$
    Many thanks.Fantastic. I really should not forget this analytic continuity issue. I have another question related to the complementary error function, could you please take a look? math.stackexchange.com/questions/3047021/…
    $endgroup$
    – gouwangzhangdong
    Dec 20 '18 at 0:43










1




1




$begingroup$
After dropping the constant factor $e^{a k}$ and differentiating wrt $k$, we need to show that $$int mathcal L_{t to s} {left[ -frac {e^{-a k - k^2/(4 t)}} {sqrt{pi t}} right]} dk = int -frac {e^{-a k - k sqrt s}} {sqrt s} dk = frac {e^{-a k - k sqrt s}} {sqrt s (sqrt s + a)}.$$
$endgroup$
– Maxim
Dec 17 '18 at 11:48






$begingroup$
After dropping the constant factor $e^{a k}$ and differentiating wrt $k$, we need to show that $$int mathcal L_{t to s} {left[ -frac {e^{-a k - k^2/(4 t)}} {sqrt{pi t}} right]} dk = int -frac {e^{-a k - k sqrt s}} {sqrt s} dk = frac {e^{-a k - k sqrt s}} {sqrt s (sqrt s + a)}.$$
$endgroup$
– Maxim
Dec 17 '18 at 11:48














$begingroup$
Thanks for that. I like your method. My way is simply doing this integral step by step. However, an initial condition is required in your way. I take $k=0$, in that case, we need a>0 for the Laplace transform of $e^{a^2 t}erfc(asqrt{t})$. I am wondering how to extend "a" to the whole complex plane.
$endgroup$
– gouwangzhangdong
Dec 19 '18 at 2:16






$begingroup$
Thanks for that. I like your method. My way is simply doing this integral step by step. However, an initial condition is required in your way. I take $k=0$, in that case, we need a>0 for the Laplace transform of $e^{a^2 t}erfc(asqrt{t})$. I am wondering how to extend "a" to the whole complex plane.
$endgroup$
– gouwangzhangdong
Dec 19 '18 at 2:16






1




1




$begingroup$
You can extend your formula to $k geq 0 land a in mathbb C$. $int_0^infty e^{a^2 t} operatorname{erfc}(a sqrt t) e^{-s t} dt$ converges for $$(operatorname{Re} a geq 0 land operatorname{Re} s > 0) lor (operatorname{Re} a < 0 land operatorname{Re} s > max(operatorname{Re}a^2, 0)),$$ in agreement with the location of the zeroes of $sqrt s (sqrt s + a)$. Another way is to multiply back by $e^{a k}$ and take $k = infty$ to show that the integration constant is zero.
$endgroup$
– Maxim
Dec 19 '18 at 14:52




$begingroup$
You can extend your formula to $k geq 0 land a in mathbb C$. $int_0^infty e^{a^2 t} operatorname{erfc}(a sqrt t) e^{-s t} dt$ converges for $$(operatorname{Re} a geq 0 land operatorname{Re} s > 0) lor (operatorname{Re} a < 0 land operatorname{Re} s > max(operatorname{Re}a^2, 0)),$$ in agreement with the location of the zeroes of $sqrt s (sqrt s + a)$. Another way is to multiply back by $e^{a k}$ and take $k = infty$ to show that the integration constant is zero.
$endgroup$
– Maxim
Dec 19 '18 at 14:52












$begingroup$
Many thanks.Fantastic. I really should not forget this analytic continuity issue. I have another question related to the complementary error function, could you please take a look? math.stackexchange.com/questions/3047021/…
$endgroup$
– gouwangzhangdong
Dec 20 '18 at 0:43






$begingroup$
Many thanks.Fantastic. I really should not forget this analytic continuity issue. I have another question related to the complementary error function, could you please take a look? math.stackexchange.com/questions/3047021/…
$endgroup$
– gouwangzhangdong
Dec 20 '18 at 0:43












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