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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?

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

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