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I have trouble in integrating the following integral.

I would appreciate any help :D

$$\int_0^1 \sqrt{-\log x}\, a\, x^{a-1}dx$$

Thanks heaps :D

The answer is $\sqrt{\pi}/2(\sqrt{a})$.

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3 Answers 3

Hint: Let $x=e^{-t^2/2}$. ${}{}{}{}{}{}{}{}{}$

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Let $x=e^{-t^2}$. We then have $-\log(x) = t^2$. Hence, we get that $$I = \int_{\infty}^0 t a e^{-(a-1)t^2}(-2t) e^{-t^2} dt = 2a \int_0^{\infty} t^2 e^{-at^2} dt$$ and this gives you the answer. If $$J(a) = \int_0^{\infty} e^{-at^2} dt$$ we then have $J(a) = \dfrac12 \sqrt{\dfrac{\pi}a}$. Hence, $$J'(a) = -\int_0^{\infty} t^2 e^{-at^2} dt = -\dfrac{I}{2a}$$ But we also have $J'(a) = -\dfrac14 \sqrt{\dfrac{\pi}{a^3}}$. This gives us $$I = \dfrac12 \sqrt{\dfrac{\pi}a}$$

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Substitute $x=e^{-t}$ and transform the integral into

$$a \int_0^{\infty} dt\: e^{-a t} \sqrt{t}$$

and express this in terms of a Gamma function by substituting $u=a t$:

$$a \frac{1}{a} a^{-1/2} \int_0^{\infty} du \: u^{1/2} e^{-u} = a^{-1/2} \Gamma{\left(\frac{3}{2} \right)} = \frac{1}{2} \sqrt{\frac{\pi}{a}}$$

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@AméricoTavares: thanks. Lots of those these days. –  Ron Gordon Apr 5 '13 at 16:04
    
Who down voted and why? –  Ron Gordon Apr 7 '13 at 14:27

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