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I've got $V(\alpha) = \int_0^{\overline{w}(\alpha)} F^\alpha(w)(1-F^\alpha(w)) dw$, where $\overline{w}(\alpha)=\frac{2\alpha\pi}{1+\alpha}$ and $F^\alpha(w) = \frac{1-\alpha}{2\alpha} \frac{w}{\pi-w} $.

I am trying to find $\lim_{\alpha \to 0} V(\alpha)$. Is there any way to do this without evaluating the integral?

Thanks so much for your help!

Best wishes, Leon

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1 Answer 1

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You have $$V(\alpha) = \int_0^{2\alpha\pi \over 1 + \alpha} \bigg({1 - \alpha \over 2\alpha}{w \over \pi - w}\bigg) \bigg(1 - {1 - \alpha \over 2\alpha} {w \over \pi - w}\bigg)dw$$ Changing variables $w = \alpha u$, this becomes $$V(\alpha) = \alpha \int_0^{2\pi \over 1 + \alpha} \bigg({1 - \alpha \over 2}{u \over \pi - \alpha u}\bigg) \bigg(1 - {1 - \alpha \over 2}{u \over \pi - \alpha u}\bigg)du$$ As $\alpha \rightarrow 0$, the integral approaches $$\int_0^{2\pi }{u \over 2\pi}\bigg(1 - {u \over 2\pi}\bigg)du$$ $$= {\pi \over 3}$$ So due to the $\alpha$ factor, $\lim_{\alpha \rightarrow 0} V(\alpha) = 0$. And you even have $V'(0) = {\pi \over 3}$.

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Thanks. I understand how to get to the integrand being $\frac{u}{2\pi}(1-\frac{u}{2\pi})$, but how do you get it being to $\frac{u}{2\pi}$ from there? –  LeonM Oct 27 '13 at 16:10
    
@user97209 You are correct... I corrected my answer and you can see it's still zero. –  Zarrax Oct 27 '13 at 16:18
    
Awesome. Yep, I see it, cheers. –  LeonM Oct 27 '13 at 16:18

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