3
$\begingroup$

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 $x$.

$\endgroup$
  • 1
    $\begingroup$ 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
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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...

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

$\endgroup$
0
$\begingroup$

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.

EDIT

Notice that in my last result we have

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

hence as $\epsilon \to 0$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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