Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

For the given function $$f(x) = \int_{x}^1 \frac{tx}{\sqrt{t^2-x^2}} \,dt$$ with -1 < x < 1.

  1. Calculate the improper integral.
  2. Calculate the Taylor series of $f(x)$ at $x=0$ until the third order.

This is a exercise in a old exam. I tried different stuff, but I did not got very far. I would really appreciate any help.

share|cite|improve this question
up vote 2 down vote accepted

Notice that the integrand is just the derivative of the denominator: $$\frac{d}{dt}\sqrt{t^2-x^2} =2t\cdot\frac{1}{2}(t^2-x^2)^{-1/2} =\frac{t}{\sqrt{t^2-x^2}}$$ So that: $$\int \frac{x t}{\sqrt{t^2-x^2}} dt = x\int \frac{t}{\sqrt{t^2-x^2}} dt= x\sqrt{t^2-x^2}$$ Therefore: $$f(x) = \left. x \sqrt{t^2-x^2} \right|^1_x = x \sqrt{1-x^2}$$

The Taylor series is a bit long, but you can start with the Taylor expansion of $g(x)=\sqrt{1-x^2},\ g(0)=1$: $$g'(x)=-\frac{x}{\sqrt{1-x^2}},\ g'(0)=0$$ $$g''(x)=-\frac{x^2}{\left(1-x^2\right)^{3/2}}-\frac{1}{\sqrt{1-x^2}}, \ g''(0)=-1$$ So that: $$g(x) = 1 -\frac{x^2}{2}+O(x^4)$$ And therefore: $$f(x) = xg(x) = x -\frac{x^3}{2} +O(x^4)$$

share|cite|improve this answer
This is strange as with the denominator $\sqrt(t^2-x^2)$ i get $\frac{d}{dt} \sqrt(t^2-x^2) = -\frac{x}{\sqrt(t^2-x^2)}$ which is not the nominator. – leo Jan 9 '13 at 14:25
@leo, I meant the integrand. – nbubis Jan 9 '13 at 14:30
ok, but why does this follow? I do not know this rule. Any hints/links? – leo Jan 9 '13 at 14:39
I think you wrote it yourself. since the derivative and the integral are in a sense opposites, then if the integrand is just the derivative of the denominator, then then the integral is just the denominator. – nbubis Jan 9 '13 at 14:44
@leo, understand it now? – nbubis Jan 9 '13 at 15:05

$$ f(x) = \int_{x}^1 \frac{tx}{\sqrt{t^2-x^2}} \,dt = \frac{x}{2}\int_{x}^1 \frac{2t}{\sqrt{t^2-x^2}}dt= {x}\sqrt{t^2-x^2}|_{t=x}^{t=1}=x\sqrt{1-x^2}. $$

To derive the Taylor series, you can use the binomial theorem

$$ f(x)= x\sqrt{1-x^2} = x\sum_{k=0}^{\infty} {1/2\choose k}(-x^2)^{k}= x(1-\frac{1}{2}x^2+\dots) = x -\frac{1}{2}x^3+\dots, $$


$$ {n\choose k}=\frac{n!}{k!(n-k)!}. $$

share|cite|improve this answer
Thank you! Can you also help me with the series? – leo Jan 9 '13 at 14:37

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.