Laplace Transform of Error Function: Laplace-Transform of (erfc(1/sqrt{t}))


Full Member
Aug 19, 2020
There are some steps that if I apply them I can show that [imath]\displaystyle \mathscr{L}\{\text{erfc}{(\sqrt{t})}\} = \frac{1}{s}\left(1 - \frac{1}{\sqrt{s + 1}}\right)[/imath]. If I apply the same steps to find [imath]\displaystyle \mathscr{L}\bigg\{\text{erfc}\bigg(\frac{1}{\sqrt{t}}\bigg)\bigg\}[/imath], they don't work.

How do I show the following?

[imath]\displaystyle \mathscr{L}\bigg\{\text{erfc}\bigg(\frac{a}{2\sqrt{t}}\bigg)\bigg\} = \dfrac{e^{-a\sqrt{s}}}{s}[/imath]

Can the first identity help me to find the second identity or I have to start everything from scratch?
I don't immediately see a way to use the first identity. I will attempt to show by definition of Laplace transform.

[math] \displaystyle\mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} = \int_0^\infty e^{-st} \left(1 - \dfrac2{\sqrt \pi} \int_0^{\frac a{2\sqrt t}} e^{-x^2}\,dx\right)\,dt = \dfrac1s - \dfrac2{\sqrt \pi} \int_0^\infty \int_0^{\frac a{2\sqrt t}} e^{-st} e^{-x^2}\,dx\,dt [/math]
We change the order of integration of the double integral:
[math] \displaystyle\mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} = \dfrac1s - \dfrac2{\sqrt \pi} \int_0^\infty \int_0^{\frac{a^2}{4x^2}} e^{-st} e^{-x^2}\,dt\,dx = \dfrac1s - \dfrac2{\sqrt \pi} \int_0^\infty \dfrac{1 - \exp\left(-\dfrac{a^2s}{4x^2}\right)}s e^{-x^2}\,dx [/math]
Expand and we have
[math] \displaystyle\mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} =\dfrac1s - \dfrac2{s\sqrt \pi} \int_0^\infty e^{-x^2}\,dx + \dfrac2{s \sqrt \pi} \int_0^\infty \exp\left(-x^2 - \dfrac{a^2s}{4x^2}\right)\,dx = \dfrac2{s \sqrt \pi} \int_0^\infty \exp\left(-x^2 - \dfrac{a^2s}{4x^2}\right)\,dx = \dfrac2{s\sqrt \pi} \int_0^\infty \exp\left(-\left(x - \dfrac{|a|\sqrt s}{2x}\right)^2 - |a| \sqrt s\right)\,dx [/math]
[math] \displaystyle \mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} = \dfrac{2e^{-|a|\sqrt s}}{s\sqrt \pi} \int_0^\infty \exp\left(-\left(x - \dfrac{|a|\sqrt s}{2x}\right)^2\right)\,dx [/math]
We are left with an integral that can be solved easily with the right trick. Use the Cauchy-Schlömilch transformation here and we get
[math] \displaystyle \mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} = \dfrac{2e^{-|a|\sqrt s}}{s\sqrt \pi} \int_0^\infty \exp\left(-x^2\right)\,dx [/math]
The rest is trivial.
I started solving this problem in the same way you did, but did not see the idea of changing the order of integration. What a bright mind you have to spot that.

The trick that you did at the end, Cauchy-Schlömilch transformation, I have never heard of it before. The funny thing was that we tried to solve an integral a few months ago, and we got the same form of this transformation, but we were stuck to solve it. We used all the ideas we knew but failed to arrive to a solution until the thread was closed.

See the thread, post #4

Even the Great Dan, did not see this trick!😱

Thank you MaxWong. Everyday, we learn something new.