$$ \mathcal{L} \{ erfc(\sqrt t) \}$$
$$ =\mathcal{L} \{ {2\over√π}\int_{√t}^∞e^{-u^2}du \}$$
$$ = {2\over√π} \int_0^∞[ \int_{√t}^∞e^{-u^2}du] e^{-st}dt$$
$$ = {2\over√π} \int_0^∞[ \int_{0}^{u^2}e^{-st}dt] e^{-u^2}du$$
$$ = {2\over√π} \int_0^∞ {e^{-st}\over-p} {|}^{u^2}_0 e^{-u^2}du$$
$$ = {2\over√π} \int_0^∞ {1-e^{-u^{2}s}\over p} e^{-u^2}du$$

Question

$$ \mathcal{L} \{ erfc(\sqrt t) \}$$
$$ =\mathcal{L} \{ {2\over√π}\int_{√t}^∞e^{-u^2}du \}$$
$$ = {2\over√π} \int_0^∞[ \int_{√t}^∞e^{-u^2}du] e^{-st}dt$$
$$ = {2\over√π} \int_0^∞[ \int_{0}^{u^2}e^{-st}dt] e^{-u^2}du$$
$$ = {2\over√π} \int_0^∞ {e^{-st}\over-p} {|}^{u^2}_0  e^{-u^2}du$$
$$ = {2\over√π} \int_0^∞ {1-e^{-u^{2}s}\over p} e^{-u^2}du$$

unklerhaukus · Answer

am i doing this right

jamesj · Answer

When you changed order of integration, I don't think you've got the limits right.

unklerhaukus · Answer

ok
|dw:1329697649791:dw|

jamesj · Answer

$$ \int_0^\infty \int_{\sqrt{t}}^\infty ... \ du \ dt = \int_0^\infty \int_{u^2}^\infty ... dt \ du $$

unklerhaukus · Answer

isnt your right hand side that this region instead? |dw:1329697843332:dw|