Closed Form of $\int_{1}^{ \infty } { \frac{1}{x}\operatorname{erfc}\left[\frac{\log x}{2 k} -\frac{k}{2}\right]{\kern 1pt} \,dx}$

Question

I am looking for a closed form of the following $$I = \int_{1}^{ \infty } { \frac{1}{x}\operatorname{erfc}\left[\frac{\log x}{2 k} -\frac{k}{2}\right]{\kern 1pt} \,dx},$$ for $k \ge 0$. Upon Integrating I found the anti-derivative $$\int { \frac{1}{x}\operatorname{erfc}\left[\frac{\log x}{2 k} -\frac{k}{2}\right]{\kern 1pt} \,dx} = \left(\log (x)-k^2\right) \operatorname{erfc}\left(\frac{\log (x)-k^2}{2 k}\right)-\frac{2 k e^{-\frac{\left(k^2-\log (x)\right)^2}{4 k^2}}}{\sqrt{\pi }}$$ but when I substitute the limit I got $\infty \times 0$ term. Is there another way to find a closed form for this integral. Thank you.

Answer 1

To compute \begin{align} \lim_{x\to\infty} (\log(x)-k^{2})\operatorname{erfc}\left(\frac{\log(x)-k^{2}} {2k}\right), \end{align} we can apply L'Hopital's rule. Let $y=\log(x)$, then \begin{align} \lim_{y\to\infty}\frac{2}{\sqrt{\pi}}\frac{\int_{\frac{y-k^{2}}{2k}}^{\infty}e^{-t^{2}}dt}{1/(y-k^{2})}=\lim_{y\to\infty}\frac{2}{\sqrt{\pi}}\frac{-\frac{1}{2k}\exp\left(-\left(\frac{y-k^{2}}{2k}\right)^{2}\right)}{-\frac{1}{(y-k^{2})^{2}}}=\frac{1}{k\sqrt{\pi}}\lim_{y\to\infty}\frac{(y-k^{2})^{2}}{\exp\left(\left(\frac{y-k^{2}}{2k}\right)^{2}\right)}=0 \end{align}