# Limit of $\int_0^a \sqrt{\frac{x^2+1}{x(a-x)}} \mathrm dx,a\rightarrow 0^+$

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?

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## 1 Answer

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! – Chon Jun 2 '12 at 18:10
What justifies the change of limit and integral? – Pedro Tamaroff Jun 4 '12 at 21:27
@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 $\forall a \leq 1$. – user17762 Jun 4 '12 at 21:48