Asymptotics for $f(\lambda)=\int_1^\infty\frac{1}{\sqrt{x}(\lambda+1)^x}dx$ as $\lambda\to0^+$. I would like to find asymptotics, as $\lambda\to0^+$ ($\lambda$ is a positive parameter), for
$$
f(\lambda)=\int_1^\infty\frac{1}{\sqrt{x}(\lambda+1)^x}dx.
$$
There is uniform convergence to $\frac{1}{\sqrt{x}}$ of the integrand, as $\lambda\to0^+$, and hence $f(\lambda)\to+\infty$, as $\lambda\to0$, but I would like to find the "velocity" of this divergence. Are there any simple estimates that would lead to the asymptotics that I am not seeing?
 A: Let $u=\sqrt x$ so \begin{align}f(\lambda)&=2\int_1^\infty\frac{dx}{(\lambda+1)^{x^2}}=\sqrt{\frac\pi{\log(\lambda+1)}}\left(1-\operatorname{erf}\sqrt{\log(\lambda+1)}\right)\end{align} where the last equality is reached by recognising that $(\lambda+1)^{-x^2}$ is akin to a Gaussian density. A Taylor series expansion gives $$f(\lambda)=\sqrt{\frac\pi{\log(\lambda+1)}}-2\sum_{n=0}^\infty\frac{(-1)^n\log^n(\lambda+1)}{n!(2n+1)}=\sqrt{\frac\pi\lambda}+\sum_{n=0}^\infty a_n\lambda^n$$ for some real $a_n$. Thus $f$ diverges at a rate of $\mathcal O(\lambda^{-1/2})$ as $\lambda\to0^+$.
A: $$\int_1^\infty\frac{1}{\sqrt{x}(\lambda+1)^x}\,dx=\int_1^\infty\frac{e^{-\ln(\lambda+1)x}}{\sqrt{x}}\,dx$$
now try using the substitution $u^2=x\Rightarrow dx=2u\,du$ and so:
$$2\int_1^\infty e^{-\ln(\lambda+1)u^2}\,du$$
now letting $v=\sqrt{\ln(\lambda+1)}u\Rightarrow du=\frac{1}{\sqrt{\ln(\lambda+1)}}dv$ which gives us:
$$\frac{2}{\sqrt{\ln(\lambda+1)}}\int\limits_{[\ln(\lambda+1)]^{1/2}}^\infty e^{-u^2}\,du=\sqrt{\frac{\pi}{\ln(\lambda+1)}}\operatorname{erfc}\left(\sqrt{\ln(\lambda+1)}\right)$$
now as $\lambda\to0^+,\sqrt{\ln(\lambda+1)}\to0$ and so $\operatorname{erfc}\left(\sqrt{\ln(\lambda+1)}\right)\to1^-$. This means that the rate of divergence of this integral will just be:
$$\sqrt{\frac{\pi}{\ln(\lambda+1)}}\,\,\,\,,\lambda\to0^+$$
