Question

Help with this integration?

\[\int^\infty_0 \text{erfc}\left(\frac{z}{2\sqrt{kt}}\right) dz\]

Answer 1

so after u substitutions and stuff I get

\[2\sqrt{\frac{kt}{\pi}}\left(e^{-\infty}-e^0\right)\]
\[=-2\sqrt{\frac{kt}{\pi}}\]

Answer 2

but the answer has no negative sign? can anyone see where I am missing a negative? It's not in the U substiution...

Answer 3

Never played with the erfc function but after the sub we should have something like this right?
\[\int\limits_{0}^{\infty} 2 \sqrt{kt } \ \cdot \erfc(u) du \\ \text{ which looks like this equals } \\ =\lim_{z \rightarrow \infty }2 \sqrt{ kt }[u \erfc(u)-\frac{e^{-u^2}}{\sqrt{\pi}}]_0^z\]

is that correct or seem right to you?

Answer 4

I have \[\int_x^\infty \text{erfc}(y)dy = \frac{e^{-x^2}}{\sqrt{\pi}}-x\text{erfc}(x)\]

Answer 5

http://mathworld.wolfram.com/Erfc.html
I was going by that indefinite integral result given here

Answer 6

ok, i don't quite get it, but I know that the answer is correct. at least acording to mathematica haha! thanks.

Answer 7

\[\int\limits\limits\limits_{0}^{\infty} 2 \sqrt{kt } \ \cdot \erfc(u) du \\ \text{ which looks like this equals } \\ =\lim_{z \rightarrow \infty }2 \sqrt{ kt }[u \erfc(u)-\frac{e^{-u^2}}{\sqrt{\pi}}]_0^z \\ =2 \sqrt{kt }\lim_{z \rightarrow \infty}((z \erfc(z)-\frac{e^{-z^2}}{\sqrt{\pi}})-(0 \erfc(0)-\frac{e^{-0^2}}{\sqrt{\pi}})) \\ =2 \sqrt{kt }(0-\frac{0}{\sqrt{\pi}}-0+\frac{1}{\sqrt{\pi}} ) \\ =2 \sqrt{kt}(\frac{1}{\sqrt{\pi}}-\frac{0}{\sqrt{\pi}}) \]

Answer 8

\[=\frac{2 \sqrt{kt}}{\sqrt{\pi}}\]

Answer 9

just plugged in my upper and lower limit
and evaluated the limit as z-> inf

Answer 10

are you cool with that?

Answer 11

thanks!

Answer 12

don't normally post my whole work
but never dealt with the erfc function before :p