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This question is similar to my previous one:

I would like to find the limit of $$ \int_0^a \sqrt{\frac{x^2+1}{x(a-x)}} \mathrm dx$$

when $$ a\rightarrow 0^+$$ Once again it seems that $$ \int_0^a \sqrt{\frac{x^2+1}{x(a-x)}} \mathrm dx\sim_{a\rightarrow 1^+} \pi$$

We have:

$$ \sqrt{\frac{x^2+1}{x(a-x)}}=\frac{2}{a}\sqrt{\frac{x^2+1}{1-(\frac{2x}{a}-1)^2}} $$

Does this help find a suitable change of variable?

1 Answers 1

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Try $x = at$. The integral becomes $$I = \int_0^1 \sqrt{\dfrac{a^2t^2+1}{at(a-at)}} a dt = \int_0^1 \sqrt{\dfrac{a^2t^2+1}{t(1-t)}} dt$$ and now taking the limit as $a \rightarrow 0$ gives us $$I = \int_0^1 \dfrac{dt}{\sqrt{t(1-t)}} = \pi$$

In general, the idea is to have the limits of the integral independent of $a$ or the integrand independent of $a$ and then take the limit as $a \rightarrow 0$.

Through substitution, it is more often easier to get the limits independent of $a$. Once you have this take the limit as $a \rightarrow 0$.

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    Ok, thank you for your clear explanation!2012-06-02
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    What justifies the change of limit and integral?2012-06-04
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    @PeterTamaroff For instance, dominated convergence theorem will be sufficient here. Let $a \to 0$ as say $\dfrac1n$. You can bound $\sqrt{\dfrac{1+a^2t^2}{t(1-t)}}$ by $\sqrt{\dfrac{2}{t(1-t)}}$ $\forall t \in [0,1]$ and $\forall a \leq 1$.2012-06-04