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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.

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    $\begingroup$ 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 $e^{a^2 t}erfc(a\sqrt{t})$. I am wondering how to extend "a" to the whole complex plane. $\endgroup$ – gouwangzhangdong Dec 19 '18 at 2:16
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    $\begingroup$ 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

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