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?

  • 2
    $\begingroup$ How about $u=\sqrt{-\ln{(x)}}\implies x=e^{-u^2}$? In these bounds the integral simplifies to just $\int_0^1\sqrt{-\ln{(x)}}\mathrm{d}x$. $\endgroup$ Jul 12, 2019 at 8:37
  • 1
    $\begingroup$ first take -lnx=t, (for lower bound, take Lim x tends to 0). Then t=s^2 followed by integration by parts $\endgroup$
    – user600016
    Jul 12, 2019 at 8:40

1 Answer 1


Let $t=\sqrt{|\ln x|}$ for $0<x<1$, then $x=e^{-t^2}$ and thus $$\int_0^1 \sqrt{|\ln x|}\,dx=-\int_0^\infty t\,d(e^{-t^2})=-te^{-t^2}|_0^\infty+\int_0^\infty e^{-t^2}\,dt=\alpha.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.