I need to calculate the next integral: $\int_{x}^{\infty }\operatorname{erfc}(\frac{\xi }{2 \sqrt{D\cdot t}})d\xi $
My attempt:
$u=\frac{\xi }{2\sqrt{D\cdot t}}\rightarrow du=\frac{d\xi }{2\sqrt{D\cdot t}}$
$2\sqrt{D\cdot t}\int \operatorname{erf}(u)du\rightarrow 2\sqrt{D\cdot t}[u\cdot \operatorname{erfc}(u)-\frac{e^{-u^{2}}}{\sqrt{\pi }}]\rightarrow 2\sqrt{D\cdot t}[\frac{\xi }{2\sqrt{D\cdot t}}\cdot erfc(\frac{\xi }{2\sqrt{D\cdot t}})-\frac{e^{-\frac{\xi ^{2}}{4Dt}}}{\sqrt{\pi }}]|_{x}^{\infty }$
At $\infty$ we will get:$\frac{\infty }{2\sqrt{Dt}}\operatorname{erfc}(\infty )-\frac{1}{e^{\frac{\infty }{4Dt}}\sqrt{\pi }}\rightarrow \infty \cdot 0$
Where is my mistake? Thanks for the help in advance
(i) Find the integral $$\int_{x}^berfc(\frac{\xi }{2 \sqrt{D\cdot t}})d\xi$$. This integral can be calculated by writing $$\int_{x}^berfc(\frac{\xi }{2 \sqrt{D\cdot t}})d\xi$$ $$=\int_{x}^b (\xi)' erfc(\frac{\xi }{2 \sqrt{D\cdot t}})d\xi$$ and integrating by parts. $$\int_{x}^b (\xi)' erfc(\frac{\xi }{2 \sqrt{D\cdot t}})d\xi$$ $$=(\xi erfc(\frac{\xi }{2 \sqrt{D\cdot t}}))|_x^b -\int_{x}^b \xi erfc'(\frac{\xi }{2 \sqrt{D\cdot t}})d\xi$$ (ii) Take the expression you found for the integral in (i) and take the limit as $b \to \infty$. Note that the integral on the right above is easy because $erfc'$ is just $constant \text { } e^{- \xi^2/constant}$ and thus can be found immediately upon substituting $\zeta = \xi^2$ The limit $$\lim_{b \to \infty}b \text { } erfc(\frac{b }{2 \sqrt{D\cdot t}})$$ $$=\lim_{b \to \infty}\frac {erfc(\frac{b }{2 \sqrt{D\cdot t}})}{\frac{1}{b}}$$ and this limit can be found by l"Hospital's Rule.