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

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?

• why do you not cancel the both $t$? – Dr. Sonnhard Graubner Jan 22 '18 at 17:27
• $$\int_{1}^{\sqrt{x}}\frac{dt}{t\sqrt{\log t}}\stackrel{t\mapsto e^z}{=}\int_{0}^{\frac{1}{2}\log x}\frac{dz}{\sqrt{z}}=\sqrt{2\log x}.$$ – Jack D'Aurizio Jan 22 '18 at 17:28
• $u=\ln(t)$ implies $du=\dfrac{1}{t}$ – JohnColtraneisJC Jan 22 '18 at 17:28
• @JackD'Aurizio thanks but I want to consider $t^2 \sqrt{\log{t}}$ in the denominator. – Khadija Mbarki Jan 22 '18 at 17:38
• I mean that the integral is $$\sqrt{\pi}\,\text{Erf}\left(\sqrt{\tfrac{1}{2}\log x}\right).$$ – Jack D'Aurizio Jan 22 '18 at 17:45

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: $$I=\sqrt{\pi}erf(\sqrt{\frac{1}{2}\log x})$$