I know the following integral should be: $$ \int_{0}^{1} \frac{dx}{\sqrt{1-x^2-\epsilon(1-x^3)}} \approx \pi/2 + \epsilon$$ for $\epsilon$ small. What I do is set $-\epsilon(1-x^3)=y$ and then since $y$ it's always greatly smaller than $(1-x^2)$ I do Taylor expansion around $y=0$: $$ \frac{dx}{\sqrt{1-x^2+y}} \approx \frac{dx}{\sqrt{1-x^2}}-\frac{y \cdot dx}{2 \cdot (1-x^3)^{3/2}} $$

Then I recover $y=-\epsilon(1-x^3)$ and I get the correct answer. The problem is that I don't know if that's mathematically correct because $y$ depends on $x$.

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    $\begingroup$ What's wrong with having $y$ depend on $x$? I mean, which step in the process do you think could be flawed for that reason? $\endgroup$ – tilper Nov 20 '17 at 18:08
  • $\begingroup$ I would worry near $x=1$. There, both $1-x^2$ and $1-x^3$ are small. $\endgroup$ – Ron Gordon Nov 20 '17 at 18:22
  • $\begingroup$ I'm worried about doing Taylor in that way isn't correct, for example for $y=0$ I have $x=1$ then $f(0) = \infty$ so I'm doing Taylor around $\infty$ and makes no sense to me $\endgroup$ – cramirpe Nov 20 '17 at 18:23

Using Taylor series you will end up having to justify evaluation at the improper bound of $1$, which will require further details that neither of the other answers have addressed. Instead, you could just note that

$$\frac{1-x^3}{1-x^2} = \frac{(1-x)(1+x+x^2)}{(1-x)(1+x)} = 1 + \frac{x^2}{1+x} \in [1,\tfrac32)$$ for $x \in [0,1)$ and hence

$$\frac{1}{\sqrt{1-x^2}} \leq \frac{1}{\sqrt{1-x^2 - \epsilon(1-x^3)}} = \frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\sqrt{1 - \epsilon \frac{1-x^3}{1-x^2}}} \leq \frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\sqrt{1 - \tfrac32\epsilon }} $$ for all $0<\epsilon <\tfrac23$.

So by integrating we get: $$ \frac{\pi}{2} \leq \int_0^1 \frac{dx}{\sqrt{1-x^2 - \epsilon(1-x^3)}} \leq \frac{\pi}{2} \cdot \frac{1}{\sqrt{1 - \tfrac32\epsilon }}$$ We may then take the series expansion about $0$ of the right hand side, giving $$\frac{1}{\sqrt{1-\tfrac32 \epsilon}} = 1 + \tfrac34 \epsilon + O(\epsilon^2)$$

Hence $$\frac{\pi}{2} \leq \int_0^1 \frac{dx}{\sqrt{1-x^2 - \epsilon(1-x^3)}} \leq \frac{\pi}{2} + \frac{3\pi}{8} \epsilon + O(\epsilon^2)$$


You are on safer ground factoring the $1-x$ out of the square root as follows:

$$\begin{align} I(\epsilon) &= \int_0^1 \frac{dx}{\sqrt{1-x^2-\epsilon (1-x^3)}} \\ &= \int_0^1 dx \frac{(1-x)^{-1/2}}{\sqrt{1+x-\epsilon(1+x+x^2)}} \\ &= \int_0^1 dx \, \frac{x^{-1/2}}{\sqrt{2-x-\epsilon (3-3 x+x^2)}} \\ &=2 \int_0^1 dx \left [ 2-x^2 - \epsilon (3-3 x^2+x^4)\right ]^{-1/2} \\ &= 2 \int_0^1 dx (2-x^2)^{-1/2} \left [1-\epsilon \frac{3-3 x^2+x^4}{2-x^2} \right ]^{-1/2} \end{align}$$

I hope you see where this is heading. You may now expand the term in brackets in a Taylor expansion in $\epsilon$ knowing that the term involving $\epsilon$ is small over the entire region of integration. Use trig substitution and the answer is straightforwardly...

$$I(\epsilon) = \frac{\pi}{2} + \epsilon + O(\epsilon^2)$$


You can be authorized in doing that because $\epsilon$ is small (we suppose small enough to give you the "license" of dong that).

Also notice that the range of the integral is $[0, 1]$, hence $x$ too, in a certain way, is small, and terms like $x^3$ are then even smaller.

There are many approaches for dealing with such an integral. Yours is one. Then you can take that function and make a Tylor series in $x$ or in $\epsilon$, obtaining respectively:

$$\star : \frac{1}{\sqrt{1-\epsilon }}+\frac{x^2}{2 (1-\epsilon )^{3/2}}-\frac{x^3 \epsilon }{2 (1-\epsilon )^{3/2}}+O\left(x^4\right)$$

$$\star : \frac{1}{\sqrt{1-x^2}}-\frac{\left(x^3-1\right) \epsilon }{2 \left(1-x^2\right)^{3/2}}+\frac{3 \left(x^3-1\right)^2 \epsilon ^2}{8 \left(1-x^2\right)^{5/2}}-\frac{5 \left(x^3-1\right)^3 \epsilon ^3}{16 \left(1-x^2\right)^{7/2}}+O\left(\epsilon ^4\right)$$

Those are clearly Taylor series of the whole function, that is, of

$$\frac{1}{\sqrt{1 - \epsilon - x^2 + \epsilon x^3}}$$

Another way is to think about the term $\epsilon x^3$ and rawly say "let's get rid of this", remaining with the easy integral

$$\int_0^1 \frac{\text{d}x}{\sqrt{1 - \epsilon - x^2}} = -\tan ^{-1}\left(\frac{\epsilon }{(-\epsilon )^{3/2}}\right)$$

Numerica other methods are available, but this wouldn't be the topic of the question.


Notice that in my last result we have

$$\frac{\epsilon}{(-\epsilon)^{3/2}} = \frac{1}{\sqrt{-\epsilon}}$$

hence as $\epsilon \to 0$

$$-\arctan\left(\frac{1}{\sqrt{-\epsilon}}\right) = \frac{\pi}{2}$$

  • $\begingroup$ That's a more intuitive way of doing it, Taylor with all dependence on $\epsilon$ (with exact the same result as mine) or $x$. But doing it only with a part of all the function (as I do) is correct? Notice for $y=0$, $x=0$ so I'm doing Taylor around $y=0$ when $f(0) = \infty$ $\endgroup$ – cramirpe Nov 20 '17 at 18:33
  • $\begingroup$ What about near $x=1$ where both terms are near zero? $\endgroup$ – Ron Gordon Nov 20 '17 at 18:37
  • $\begingroup$ Sorry, for $y=0$ then $x=1$ $\endgroup$ – cramirpe Nov 20 '17 at 18:39
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    $\begingroup$ @Henry Your second approach is incorrect. Note that $1-x^2-\epsilon < 0$ for $x$ close enough to $1$, so the integral is undefined. You would have to define the upper bound of integration in terms of $\epsilon$, so the integrals won't be immediately comparable. $\endgroup$ – adfriedman Nov 20 '17 at 21:22
  • $\begingroup$ Also $\epsilon$ being small isn't sufficient on its own to allow the Taylor series to be expanded under the integral. In my answer I show that the integral is bounded, so a dominated convergence theorem might be sufficient for justification. $\endgroup$ – adfriedman Nov 20 '17 at 21:22

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