Taylor series of an improper integral

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oignonsauce

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I try to calculate the first terms of the Taylor series of the following improper integral

[math]\int_\delta ^{1-\delta} dx \frac{x (1-x)}{\sqrt{x^2-\delta^2}\sqrt{(1-x)^2-\delta^2}}[/math] when [imath]\delta[/imath] is close to 0 and to order 2 in [imath]\delta[/imath].
After numerical study I think the answer is [imath]1-\delta^2[/imath] but I don't know how to prove it. I naively tried to develop the integrand to order 2 and then integrate, but I find [imath]1-\delta[/imath] instead of [imath]1-\delta^2[/imath].
If you have a method I'd be interested.
 
If we use [imath]G(\delta[/imath] for the expression in your post then, if I understand correctly, you need to compute [imath]G^\prime(0)[/imath] and [imath]G^{\prime\prime}(0)[/imath] -- correct?
Can you give us some context? I.e., what are you studying? Where does this problem come from? And what have you tried.

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blamocur said:
if I understand correctly, you need to compute [imath]G^\prime(0)[/imath] and [imath]G^{\prime\prime}(0)[/imath] -- correct?
Click to expand...
Yes more or less. What I want is find an asymptotic expansion for [imath] G(\delta) \begin{subarray}{c}\delta \rightarrow 0 \\\sim\end{subarray} \ 1 + f(\delta) [/imath]. In this case I want to prove that [imath] f(\delta) = -\delta^2 + ... [/imath].

what are you studying? And what have you tried
Click to expand...

This integral appeared in a physics problem, and I'd like to study the regime where [imath]\delta [/imath] is small.
I first tried to simply write the integrand to second ordre in [imath]\delta [/imath], then integrate. I got [imath]G(\delta) \sim 1- \delta [/imath], which is wrong.
I also tried to get ride of the singularities by a change of variable, but in the end I couldn't find the right result.
 
Yes I need to use the Leibniz integral rule. The difficulty is that I have to integrate the derivative of the integrand, which seems rather complicated.
 
oignonsauce said:
Yes I need to use the Leibniz integral rule. The difficulty is that I have to integrate the derivative of the integrand, which seems rather complicated.
Click to expand...
It's always easier to take a derivative than an integral. But the first order correction isn't too bad.

As to your implied question: Look up the Fundamental Theorem of Integral Calculus. You don't have take any derivatives or integrals explicitly to first order.

For the record, I've done this one a couple of ways using W|A and I get different answers. One of them was by Taylor series, like the question asks, and the others were approximations inside the integral. All of them are of the form [imath]1 + a \delta[/imath], though. (a varies depending on how I do it.)

-Dan
 
Only an idea: The integral can be written as [math]\int_\delta^{1-\delta} \frac{dx}{\sqrt{1-\alpha}\sqrt{1-\beta}} [/math] where [imath] \alpha=\dfrac{\delta^2}{x^2} \, , \,\beta=\dfrac{\delta^2}{(1-x)^2}.[/imath] With the mean value theorem for integrals, we get
[math] \int_\delta^{1-\delta} \frac{dx}{\sqrt{1-\alpha}\sqrt{1-\beta}}=\dfrac{1}{1-\dfrac{\delta^2}{(1-\xi)^2}} \int_\delta^{1-\delta} \frac{dx}{\sqrt{1-\alpha(x)}}[/math]and WA gives us the Taylor expansion for the integral:

integral (1/(root(1-y(x)))) dx= - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.
www.wolframalpha.com www.wolframalpha.com

This should at least sort out the powers of [imath] \delta [/imath].
 
Correction. I forgot a square root:

... With the mean value theorem for integrals, we get
[math] \int_\delta^{1-\delta} \frac{dx}{\sqrt{1-\alpha}\sqrt{1-\beta}}=\dfrac{1}{\sqrt{1-\dfrac{\delta^2}{(1-\xi)^2}}} \int_\delta^{1-\delta} \frac{dx}{\sqrt{1-\alpha(x)}}[/math]
 
Thanks fresh_42, this is interesting. The problem is that I am pretty sure that [imath]\xi[/imath] is a function of [imath]\delta[/imath]. And getting the [imath]\delta[/imath] dependance of [imath]\xi (\delta)[/imath] seems as hard as solving the original integral.

As to your implied question: Look up the Fundamental Theorem of Integral Calculus. You don't have take any derivatives or integrals explicitly to first order.
Click to expand...

Sorry I don't follow. I sure need to calculate the derivate of the integrand wrt [imath]\delta[/imath] and then integrate this to get [imath]G'(\delta)[/imath] (because the integrand is explicitly dependant on [imath]\delta[/imath]) no ?

All of them are of the form [imath]1+a\delta[/imath], though.
Click to expand...

Yes I also find results like this but some numerical integrations tell me that this is really wrong and the real form is [imath]1-\delta^2[/imath].
For example, for [imath]\delta=0.05[/imath], the numerical result for the integral is [imath]\simeq 0.9975[/imath] and [imath]1-\delta^2= 0.9975[/imath] whereas [imath]1-\delta =0.95[/imath]

Edit : see the image. In blue I calculated the full integral. In orange this is simply [imath]1-\delta^2[/imath].int.png
 
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blamocur said:
You are right. I forgot about the third term in the Leibniz rule and thought it is simpler than it actually is :(
Click to expand...
To make amends I ran these expressions through SymPy, which came up with the following (I switched [imath]\delta[/imath] to [imath]u[/imath] below) :
[math]g(x,u) = \frac{x \left(1 - x\right)}{\sqrt{- u^{2} + x^{2}} \sqrt{- u^{2} + \left(1 - x\right)^{2}}}[/math][math]\frac{\partial g}{\partial u} = \frac{u x \left(x - 1\right) \left(2 u^{2} - x^{2} - \left(1 - x\right)^{2}\right)}{\left(- u^{2} + x^{2}\right)^{\frac{3}{2}} \left(- u^{2} + \left(1 - x\right)^{2}\right)^{\frac{3}{2}}}[/math][math]\frac{\partial g}{\partial u}(0) = ?[/math]
But I can't even start thinking about the second derivative :)
 
Thanks for that. Nonetheless we can't take this derivative at the value [imath]u=0[/imath] (as the last line seems to imply ?). We have to integrate [imath]\partial g / \partial u[/imath] wrt [imath]x[/imath], then take [imath]u=0[/imath]. Moreover the integral of this derivative must give a divergent contribution as the first Leibniz terms are divergent. This is why I didn't try to calculate those [imath]G^{(n)}[/imath] directly, seems really hard.
 
oignonsauce said:
Nonetheless we can't take this derivative at the value u=0u=0u=0 (as the last line seems to imply ?). We have to integrate ∂g/∂u\partial g / \partial u∂g/∂u wrt xxx, then take u=0u=0u=0.
Click to expand...
Why? I don't remember all the nuances, but in most "regular" cases we can start by assigning 0 to [imath]u[/imath], and this case looks rather regular to me.
oignonsauce said:
Moreover the integral of this derivative must give a divergent contribution as the first Leibniz terms are divergent.
Click to expand...
I might be missing something, but I don't see any divergence there. As far as I can tell, [imath]g(x,0) \equiv 0[/imath], isn't it?
 
blamocur said:
Why? I don't remember all the nuances, but in most "regular" cases we can start by assigning 0 to [imath]u[/imath], and this case looks rather regular to me.
Click to expand...

I don't know. It works for [imath] G'(0) [/imath] because [imath] g'(0) =0 [/imath] but doesn't for the second order derivative as the integrals for [imath] g''(u)[/imath] are always divergent, whatever the order of the limit [imath] u \rightarrow 0 [/imath] (I verified this with Mathematica).
blamocur said:
I might be missing something, but I don't see any divergence there. As far as I can tell, [imath]g(x,0) \equiv 0[/imath], isn't it?
Click to expand...
When you apply Leibniz rule, the integrand will take the value of the integration bounds : [imath]g(x=\delta)[/imath] and [imath]g(x=1-\delta)[/imath] which are divergent. The limit [imath] \lim_{\delta \rightarrow 0 }g(x=\delta)[/imath] seems ill defined and I don't even know how I could get higher term derivatives from those, as there are infinite. For training I try to get the derivative of

[math] F(\delta) = \int_\delta^1 \frac{x}{\sqrt{x^2-\delta^2}}. [/math] I found

[math] \partial_\delta F(\delta) = -\infty - \frac{\delta}{\sqrt{1-\delta^2}} +\infty =- \frac{\delta}{\sqrt{1-\delta^2}} [/math] where one infinity comes from the first term of the Leibnez derivative rule, and the second comes from the integration of the derivative of the integral and the two cancel each other. Then I have all [imath] \partial^n_\delta F(\delta)[/imath]. But if I put [imath]\delta = 0 [/imath] directly before integrating I have problem for higher orders.

fresh_42 said:
I have taken the integral (without limits) to WA and the result was horrible despite the high degree of symmetry.

integral ((x(1-x))/(root(x^2-a^2) * root ((1-x)^2-a^2))) dx = - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.
www.wolframalpha.com www.wolframalpha.com
Click to expand...
Nice analytical result :)
 
oignonsauce said:
When you apply Leibniz rule, the integrand will take the value of the integration bounds : g(x=δ)g(x=\delta)g(x=δ) and g(x=1−δ)g(x=1-\delta)g(x=1−δ) which are divergent.
Click to expand...
You are right, and I was indeed missing something. I'd say that the two first terms of Leibniz rule are undefined since, for example, [imath]g(u,u)[/imath] (in my notation) has 0 in the denominator.
 
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