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$$\int^1_0\sqrt{\ln(1/x)}dx$$

How does one integrate the above by making use of a double integral, the only thing I could do is simplify $\ln(1/x)$ into $-\ln(x)$ which does not help much. How does one attack problems of such style?

Hagen von Eitzen
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ZF Lim
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1 Answers1

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We want $$I = \int_0^1 \sqrt{-\ln(x)} dx$$ The first thing is to get rid of this relatively ugly integrand. Let us do the most obvious substitution i.e. let $-\ln(x) = t^2$ i.e. $x = \exp(-t^2)$.

This gives us $dx = -2t \exp(-t^2) dt$. Hence, we get that $$I = \int_{\infty}^{0} t (-2t \exp(-t^2)) dt = 2 \int_0^{\infty} t^2 \exp(-t^2) dt$$ Now you should be able to evaluate this integral using the fact that $$\int_0^{\infty} \exp(- \alpha t^2) dt = \dfrac{\sqrt{\pi}}{2 \sqrt{\alpha}}$$ Look here for various proofs for the result stated in the above line.

EDIT

You could rewrite your $I$ as $$I = \int_0^1 \int_0^{-\ln(x)} \dfrac{dt dx}{2 \sqrt{t}} = \int_0^{\infty} \int_{0}^{\exp(-t)} \dfrac{dx dt}{2 \sqrt{t}} = \int_0^{\infty} \dfrac{\exp(-t)}{2 \sqrt{t}} dt = \dfrac{\sqrt{\pi}}2$$

In general, any substitution/change of variable can be interpreted (or) made into an appropriate double integral.

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  • Thanks! that method works. However, the question specifically required a method that uses the double integral more directly (although evaluating the gaussian does employ using a double integral + change of variable). Is it possible to apply double integration directly? – ZF Lim Nov 18 '12 at 10:24