Integral $\int_{1}^{\sqrt{x}} \frac{1}{t^2 \sqrt{\log{t}}}dt.$

Question

Let $x$ be a real number greater or equal to $3$. I want to compute this integral $$\int_{1}^{\sqrt{x}} \frac{1}{t^2 \sqrt{\log{t}}}dt.$$ Any help please?

Answer 1

Using $t=e^u$ we have:$$\int_{1}^{\sqrt{x}} \frac{1}{t^2 \sqrt{\log{t}}}dt=\int_{0}^{\frac{1}{2}\log x} {\frac{e^{-u}}{\sqrt u}}du$$ and $u=w^2$ gives us:$$I=\int_{0}^{\sqrt{\frac{1}{2}\log x}} 2e^{-w^2}dw$$ The latter integral has no closed form so we can describe it using error function: $$I=\sqrt{\pi}erf(\sqrt{\frac{1}{2}\log x})$$