Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 5 down vote accepted

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 $a \rightarrow 0$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.