Calculate the improper integral and the taylor series of $f(x) = \int_{x}^1 \frac{tx}{\sqrt{t^2-x^2}} \,dt$
For the given function $$f(x) = \int_{x}^1 \frac{tx}{\sqrt{t^2-x^2}} \,dt$$ with -1 < x < 1.
- Calculate the improper integral.
- 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.
$\endgroup$2 Answers
$\begingroup$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: $$f(x) = xg(x) = x -\frac{x^3}{2} +O(x^4)$$
$\endgroup$ 10 $\begingroup$$$ f(x) = \int_{x}^1 \frac{tx}{\sqrt{t^2-x^2}} \,dt = \frac{x}{2}\int_{x}^1 \frac{2t}{\sqrt{t^2-x^2}}dt= {x}\sqrt{t^2-x^2}|_{t=x}^{t=1}=x\sqrt{1-x^2}. $$
To derive the Taylor series, you can use the binomial theorem
$$ f(x)= x\sqrt{1-x^2} = x\sum_{k=0}^{\infty} {1/2\choose k}(-x^2)^{k}= x(1-\frac{1}{2}x^2+\dots) = x -\frac{1}{2}x^3+\dots, $$
where
$$ {n\choose k}=\frac{n!}{k!(n-k)!}. $$
$\endgroup$ 1
