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We are given that $$\int_{0}^\infty e^{-x^2}\,dx=\alpha$$ then we need to find the value of $$\int_{0}^{1}\sqrt{|\ln x|}\,dx$$ in terms of $\alpha$.

What I did was to integrate the second integral by parts taking $1$ as the first function and $\sqrt{|\ln x|}$ as the second.

This gave me $$\big[\sqrt{\ln x}+x\big]_{0}^{1}-\frac12\int\frac{1}{\sqrt{\ln x}}\,dx$$

But I'm stuck after this. I can't see what I should do to bring the above integral into play. Can someone suggest something to move ahead or maybe some other method?

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    $\begingroup$ How about $u=\sqrt{-\ln{(x)}}\implies x=e^{-u^2}$? In these bounds the integral simplifies to just $\int_0^1\sqrt{-\ln{(x)}}\mathrm{d}x$. $\endgroup$ – Peter Foreman Jul 12 at 8:37
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    $\begingroup$ first take -lnx=t, (for lower bound, take Lim x tends to 0). Then t=s^2 followed by integration by parts $\endgroup$ – thewitness Jul 12 at 8:40
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Let $t=\sqrt{|\ln x|}$ for $0<x<1$, then $x=e^{-t^2}$ and thus $$\int_0^1 \sqrt{|\ln x|}\,dx=-\int_0^\infty t\,d(e^{-t^2})=-te^{-t^2}|_0^\infty+\int_0^\infty e^{-t^2}\,dt=\alpha.$$

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